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ft  D I 


Y33/S- 


i.JIRY 


PLANE 
TRIGONOMETRY 


\ 


BY 

ARNOLD   DRESDEN,   Ph.D. 

Professor  of  Mathematics,  Swarthmore  College 


SECOND  PRINTING,   CORRECTED 


NEW  YORK 

JOHN  WILEY  &  SONS,  Inc. 

London:    CHAPMAN   &  HALL,   Limited 


«    ■: 


Copyright,  1921 

By 

ARNOLD  DRESDEN 


Printed  in  U.  S.  A. 


tbchna  '  i  a )mpo8ition  oo. 

CAMBRIDGE,    MASS.,     U.   S.    A.  4— 1 1 


raO- 

_     -1       v 


PREFACE 


While  the  importance  of  the  function  concept  for  elementary 
mathematics  has  become  recognized  by  many  writers  of  college 
algebra  texts  and  of  "unified  freshman  mathematics  "  books,  it 
has  received  little  recognition  from  writers  on  elementary  trigo- 
nometry. To  emphasize  this  importance  has  been  the  leading 
motive  in  writing  the  present  book.  A  somewhat  detailed  study  of 
the  graphs  of  the  trigonometric  functions  (Chapter  V)  and  of 
the  inverse  functions  (Chapter  VIII)  has  been  introduced  for 
this  purpose.  Much  more  could  and  should  be  done  in  this  di- 
rection; perhaps  the  present  effort  may  suffice  as  a  first  step. 

The  opportunity  afforded  by  the  writing  of  a  new  text  has  been 
used  to  make  some  changes  in  the  presentation  of  the  traditional 
material.  Circular  measurement  of  angles  is  introduced  in  the 
first  chapter  so  as  to  be  available  for  use  throughout  the  course. 
The  fundamental  theorems  on  projections  are  presented  early  and 
are  used  subsequently  so  that  the  student  may  be  familiar  with 
them  when  they  are  applied  in  a  general  proof  of  the  addition 
theorems,  based  on  a  method  quite  generally  followed  by  conti- 
nental writers.  Recognizing  the  value  of  the  "solution  of  tri- 
angle-," a  good  deal  of  space  has  been  devoted  to  this  subject, 
and  an  attempt  has  been  made  to  develop  it  in  such  a  manner 
that  the  student  can  appreciate  the  reasons  for  the  different 
methods  that  are  discussed. 

On  the  question  of  "  applied  problems,"  I  have  taken  a  definite 
position.  I  do  not  think  it  feasible  to  introduce  into  an  ele- 
mentary text  technical  material  from  the  applied  sciences,  impor- 
tant though  such  material  may  be.  Without  such  material,  how- 
ever, applications  cannot  well  be  anything  but  problems  which 
use  the  language  of  the  applied  sciences  without  really  belonging 
to  them.  An  elementary  text  can  render  useful  service,  even  to 
applied  science,  by  stressing  the  fundamental  concepts  of  trigonom- 
etry and  by  setting  problems  which  connect  with  the  student's 
actual  experience  and  which  suggest  ways  in  which  these  concepts 


IV  PREFACE 

may  be  applied,  leaving  actual  applications  to  the  fields  to  which 
they  belong. 

It  has  not  seemed  desirable  to  add  to  the  number  of  tables  of 
logarithms  already  available.  The  elementary  treatment  of  log- 
arithms in  Chapter  III  and  the  problems  scattered  throughout 
the  book  call  for  the  use  of  a  set  of  five-place  tables,  of  which 
there  are  many  excellent  ones  in  existence. 

No  attempt  at  logical  completeness  has  been  made,  but  rather 
has  it  been  my  aim  to  adapt  the  treatment  to  the  stage  of  logical 
development  which  may  be  expected  of  students  who  begin  the 
study  of  trigonometry.  I  am  aware  of  the  fact  that  a  fuller  dis- 
cussion might  be  made  in  several  instances  and  I  shall  be  happy 
if  the  treatment  as  given  should  arouse  the  critical  powers  of  some 
students  and  develop  in  them  a  desire  for  more  penetrating  analysis. 

The  material  as  here  presented  was  used  originally  in  mimeo- 
graphed form  by  a  few  classes  in  the  University  of  Wisconsin. 
I  am  under  a  debt  of  gratitude  to  the  Department  of  Mathematics 
for  allowing  the  material  to  be  thus  tried  out.  And  it  is  with 
special  pleasure  that  I  recognize  my  indebtedness  to  colleagues  in 
that  department  and  to  Professor  T.  M.  Simpson,  now  of  the 
University  of  Florida,  to  some  for  suggestions  and  criticisms,  to 
others  for  assistance  in  the  reading  of  the  proofs.  If  I  add  to  this 
my  appreciation  of  the  courtesy  shown  by  the  publishers  of  the 
book,  I  am  ready  to  rest  my  case  with  the  jury  consisting  of  the 
teachers  and  students  of  trigonometry. 

ARNOLD   DRESDEN 

University  of  Wisconsin 
March,  1921 


CONTENTS 


CHAPTER  I 

POSITIVE  AND   NEGATIVE  LINES  AND  ANGLES.    COORDINATES. 
RADIAN  MEASUREMENT 
Art.  Page 

1.  Directed  magnitudes     1 

2.  Points  on  a  line.     Directed  lines 1 

3,  4.   Points  in  a  plane      2 

5.  A  fundamental  theorem     3 

6,  7.  Projections  of  line  segments     4 

8,  9.  Directed  angles     5 

10,  11.  Radian  measure     6 


CHAPTER' II 

THE  TRIGONOMETRIC  RATIOS.     SIMPLE  IDENTITIES 

12,  13.  Standard  position     9 

14-17.   Trigonometric  ratios     9 

18,  19.   Reduction  to  the  first  quadrant     11 

20.   Ratios  of  acute  angles     13 

21,  22.   Ratios  of  30  °,  60  °,  and  45  °     13 

23.   Ratios  of  90°,  180°,  270°,  and  360°     15 

24,  25.    Trigonometric  functions     16 

26.    Periodicity     17 

27,  28.   Relations  between  the  trigonometric  functions     18 

CHAPTER   III 
LOGARITHMS 

29.  Theory  of  exponents     20 

30.  The  use  of  exponents  in  calculation     21 

31,  32.    Definition  of  logarithms      21 

33,  34.    Fundamental  theorems  on  logarithms     22 

35.    Common  logarithms     24 

36,  37.   Use  of  a  table  of  logarithms     26 

38,  39.    Calculation  by  means  of  logarithms     2S 

v 


VI  CONTENTS 

CHAPTER  IV 

SOLUTION  OF  RIGHT  TRIANGLES.    APPLICATIONS 

Art.  Page 

40,  41.    The  right  triangle     32 

42.  Accuracy  of  the  calculation.    Checking  the  results     33 

43,  44.   Isosceles  triangles     34 

45-47.   Projection     36 

48,  49.   Applications     38 

CHAPTER  V 
THE  GRAPHS  OF  THE  TRIGONOMETRIC  FUNCTIONS 

50.   Graphs  of  sin  0  and  cos  6     41 

51,  52.   Examples  of  graphs     42 

53,  54.   Operations  on  graphs     45 

55, 56.  Applications     47 

57,  58.   Graphs  of  tan  6  and  cot  6      48 

59.   Graphs  of  tan  (a6  +  b)  and  cot  (ad  +  b)      50 

60,  61.   Applications     51 

62,  63.   Graphs  of  sec  6  and  cosec  0     52 

CHAPTER  VI 
THE  ADDITION  FORMULAE 

64.   A  special  case      55 

65-67.   Addition  and  subtraction  formulae  for  the  sine  and  the  cosine     .  55 

68,  69.    Addition  formulae  for  the  tangent  and  the  cotangent     58 

70,  71.   Double  angle  and  half-angle  formulae     59 

72,  73.   Factorization  formulae     61 

74.  Summary     63 

75.  Exercises  on  Chapter  VI     63 

CHAPTER  VII 
THE  SOLUTION  OF  TRIANGLES 

76.  The  Law  of  Sines.     The  area  of  a  triangle 65 

77,  78.   Two  angles  and  one  side     66 

79,  80.    Two  sides  and  an  angle  opposite  one  of  them     68 

81,  82.   The  Law  of  Cosines     73 

83.  Summary  and  critique     76 

84.  The  Law  of  Tangents      76 

85,  86.    Mollweide's  [equations      78 

K7,  88.   Two  sides  and  the  included  angle     79 

89.   The  half-angle  formulae  for  the  angles  of  a  triangle     80 


CONTENTS  Vll 

Art.  Paee 

90,  91.   Threesides     81 

92.  Inscribed  and  circumscribed  circles.     Area     82 

93.  Summary     84 

94,  95.   Exercises  en  Chapter  VII       86 

CHAPTER  VIII 

INVERSE  TRIGONOMETRIC  FUNCTIONS.    TRIGONOMETRIC 
EQUATIONS 

96.    Inverse  functions     90 

97,  98.   The  graphs  of  a  pair  of  inverse  functions     92 

99.   The  inverse  sine  function     94 

100.  The  other  inverse  trigonometric  functions     95 

101.  Exercises      97 

102.  Relations  between  multiple-valued  and  single-valued  inverse 

functions     98 

103-106.   Trigonometric   equations     98 

List  of  Answers  to  the  Exercises     105 

Index    109 


PLANE    TRIGONOMETRY 


CHAPTER   I 

POSITIVE    AND    NEGATIVE    LINES    AND    ANGLES.     CO- 
ORDINATES.    RADIAN   MEASUREMENT 

1.  Directed  magnitudes.  The  use  of  positive  and  negative 
numbers  is  a  familiar  method  of  indicating  temperatures  either 
above  or  below  a  certain  fixed  temperature,  which  is  taken  as  the 
zero  of  the  scale.  In  the  same  way,  it  is  convenient  to  use  posi- 
tive and  negative  numbers  to  designate  the  measures  of  other 
magnitudes,  the  values  of  which  may  fall  on  either  side  of  a 
certain  fixed  value,  taken  as  a  point  of  departure.  For  example, 
northern  latitude  may  be  designated  as  positive,  southern  lati- 
tude as  negative;  eastern  longitude  as  positive,  western  longi- 
tude as  negative;  a  credit  balance  as  positive,  a  debit  balance  as 
negative;  altitude  above  sea  level  as  positive,  altitude  below  sea 
level  as  negative,  etc. 

2.  Points  on  a  line.  Directed  lines.  A  simple  graphical  rep- 
resentation of  such  magnitudes  is  obtained  by  means  of  a  straight 
line,  called  the  axis,  upon  which  are  indicated  a  fixed  point, 
called  the  origin,  a  unit  distance  and  a  positive  direction. 

O  (Origin)  1  Positive   Direction 

1 1 . >- 

U Unit >i  Axis 

Fig.  1 

For  every  number,  positive  or  negative,  there  is  a  single  cor- 
responding point,  P,  on  this  line.  P  is  the  point  whose  distance 
from  the  origin,  0,  measured  in  terms  of  the  unit  distance,  is,  in 
magnitude  and  direction,  equal  to  the  given  number.  Conversely, 
to  every  point,  P,  on  the  line  there  corresponds  a  real  number, 
viz..  the  measurement  of  the  distance  OP  in  terms  of  the  unit 
distance,  prefixed  by  a  plus  or  minus  sign.    With  the  conventions 

1 


2  POINTS   IN   A   PLANE 

of  Figure  1,  positive  numbers  will  correspond  to  points  to  the 
right  of  0,  negative  numbers  to  points  to  the  left  of  0.  The 
number  which  is  in  this  way  associated  with  a  point  P  is  called 
the  coordinate  of  P. 

3.  Points  in  a  plane.  A  simple  extension  of  this  method 
enables  us  to  designate  the  position  of  a  point  in  a  plane  by  means 
of  a  pair  of  numbers.  We  take  two  mutually  perpendicular 
lines,  called  the  axes  of  coordinates,  one  horizontal  and  one  verti- 
cal. Their  point  of  intersection,  called  the  origin  of  coordinates, 
is  used  as  origin  on  each  axis,  as  explained  in  the  preceding  section. 
Moreover,  a  unit  distance  and  a  positive  direction  are  specified 
on  each  axis.  The  horizontal  axis  is  called  the  axis  of  abscissae 
or  the  X-axis ;  the  vertical  line  the  axis  of  ordinates  or  the  Y-axis. 
The  position  of  an  arbitrary  point  P  in  the  plane  is  now  deter- 
mined by  the  horizontal  distance  from  the  F-axis  to  P,  meas- 
ured according  to  the  unit  and  positive  direction  on  the  X-axis, 
and  by  the  vertical  distance  from  the  X-axis  to  P,  measured  in 
accordance  with  the  specifications  on  the  F-axis.  In  this  way 
two  real  numbers  are  obtained,  called  respectively,  the  abscissa 

or  ^-coordinate  (or  simply  the 
j  "*")    and    the    ordinate    or 

y-coordinate   (or  simply  the 

-i "jr")  of  the  point   P.     The 

position  of  the  point  P  is  in- 
dicated   by    means   of   these 
~*x    two  numbers  in  parentheses, 

I the  ^-coordinate  being  placed 

first,    and    the    ^/-coordinate 
IV  second.    Thus  the  points  A,  B 

p      2  and  C  of  Figure  2  are  repre- 

sented by  the  symbols  (3,5) 
(—3,-2)  and  (2,-3)  respectively.  Conversely,  for  every  pair 
of  real  numbers,  such  as  (—  2,4),  a  point  is  uniquely  determined 
of  which  these  numbers  are  the  x-  and  ^-coordinates. 

The  four  parts  into  which  the  two  axes  of  coordinates  divide 
the  plane  are  called  the  four  quadrants;  they  are  numbered  as 
indicated  in  Figure  2.  We  say  that  A  lies  in  the  first  quadrant 
(in  I),  B  in  the  third  quadrant  (in  III)  and  C  in  the  fourth  quad- 
rant (in  IV). 


// 


111 


A  FUNDAMENTAL  THEOREM  3 

4.  Exercises. 

1.  Locate  the  points  A  (0,0),  AT  (3,5),   Y  (-4,1),  Z  (0,3),   V  (-2,  -2), 
W  (5,-2),  U  (-2,0),  T  (4,-3). 

2.  Determine  the  coordinates  of  the  third  vertex  of  an  equilateral  triangle 
of  which  the  origin  and  the  point  (6,0)  are  the  other  two  vertices. 

3.  Determine  the  coordinates  of  the  center  of  the  circle  which  passes 
through  the  points  (0,0),  (4,0),  and  (0,-4). 

4.  Show,  by  construction,  that  the  points  (0,2),  (1,5),  (3,11)  and  (  —  2,-4) 
lie  on  a  straight  line. 

5.  Show,  by  construction,  that  the  points  (—1,5),  (6,4),  (—1,-3)  and 
(6,-2)  lie  on  the  circumference  of  a  circle  of  which  the  point  (2,1)  is  the  center. 

6.  Determine  the  coordinates  of  the  center  of  the  circle  which  passes 
through  the  points  A  (5,0),  B  (0,-5)  and  C  (-5,0). 

7.  Determine  the  coordinates  of  the  points  midway  between  the  points 
A  (2,7)  and  B  (4,3);  P  (3,-4)  and  Q  (7,4);   X  (-1,3)  and  Y  (3,-1). 

8.  Find  the  distance  from  the  origin  of  each  of  the  following  points: 
A  (3,-4),  B  (-5,4),  C  (-4,-6),  D  (5,5). 

9.  If  r  denote  the  distance  of  a  point  from  the  origin,  determine  both 
coordinates  for  each  of  the  following  points: 

A:  r  =  5,  y  =  3;    B:  r  =  13,  x  =  -12;    C:  r  =  3,  y  =  -4;    D:  r  =  2,  x  =  1. 

10.  The  point  P  lies  on  a  circle  about  the  origin  as  a  center  and  with  a 
radius  equal  to  10;  its  ordinate  is  twice  its  abscissa,  the  two  coordinates  hav- 
ing like  signs.     Determine  the  coordinates  of  P. 

5.  A  fundamental  theorem. 

Theorem  I.  If  three  points  A,  B  and  C  be  taken  arbitrarily  on  a 
directed  straight  line,  the  sum  of  the  three  segments  AB,  BC  and 
CA  is  equal  to  zero. 

Proof.  Reading  from  left  to  right,  the  three  points  must  lie 
in  one  of  the  following  six  orders :  A   B   C-    A    C  B- 

b        a  c       }  B,A,C;    B,C,A; 

Fig.  3  C,  A,  B;    C,  B,  A. 

In  the  third  case,  illustrated  in  Figure  3,  the  segments  BA,  BC 
and  AC  are  positive  segments,  satisfying  the  relation  BA-\-  AC 

=  BC,  or   -BA  +BC  -AC  =  0.     But,  -BA  =  AB  and   -AC 

=  CA.     We  conclude  therefore 

,4  B  +  BC  +  CA  =  0. 

In  all  the  other  cases  the  proof  proceeds  in  entirely  analogous 
manner:  the  details  are  left  to  the  reader. 


4 


PROJECTIONS  OF  LINE  SEGMENTS 


Corollary.  If  any  number  of  points  be  taken  arbitrarily  on  a 
directed  straight  line,  the  sum  of  the  segments  from  the  first  point 
to  the  second  point,  from  the  second  point  to  the  third  point,  etc., 
and  from  the  last  point  to  the  first  point,  is  equal  to  zero. 


Proof. 
then 


Let  the  points  be  A,  B,  C,  D,  E  and  F.     We  know 


AB  +  BC  +  CA  =  0, 
EF  +  FA  +  AE  =  0, 


and 


CD  +  DE  +  EC  =  0, 

CA  +  AE  +  EC  =  0. 


If  we  now  add  the  first  three  of  these  equalities  and  subtract  the 
last  one  from  their  sum,  we  obtain 

AB  +  BC  +  CD  +  BE  +  EF  +  FA  =  0, 

which  proves  the  corollary. 

6.  Projections  of  line  segments.  In  elementary  geometry, 
the  projection  of  the  segment  AB  of  a  line  I  upon  a  line  m  is  de- 
fined as  the  segment  A-JBi  of  the  line  m  determined  by  the  feet 
Ai  and  Bx  of  the  perpendiculars  dropped  from  A  and  B  respec- 
tively upon  m.     We  prove  now  the  following  important  theorem. 

Theorem  II.  The  sum  of  the  projections  upon  a  directed  line  m  of 
the  directed  segments  of  a  closed  broken  line  is  equal  to  zero. 


Proof.  Let  the  brok- 
en line  be  ABCDEFA 
(see  Fig.  4).  The  pro- 
jections of  the  directed 
i  segments  AB,  BC,  CD, 
DE,  EF  and  FA  up- 
on the  directed  line 
m  are  the  directed 
segments  AiB\,   Bid, 


Fig.  4 

CiDi,   D1EU   ExFlt   and   FAX 
Theorem  I, 

ArBt  +  B,CX  +  CiD,  +  /),#!  +  tfiF,  +  FiAi 

hence  the  theorem  is  proved. 


In   virtue   of    the   Corollary   to 

0; 


Corollary.  The  sum  of  the  projections  upon  a  directed  line  m 
of  the  directed  segments  of  an  open  broken  line  ABC  .  .  .  XP  is  equal 
to  the  projection  upon  m  of  the  directed  line  AP. 


DIRECTED   ANGLES  5 

Proof.  If  the  projections  of  A,  B,  C, ...  X,  P  upon  m  be  denoted 
by  Ah  Bi>  d,  .  .  .  Xi,  Pi,  we  have  to  show  that 

A1B1  +  B1C1  +  •  •  •  +  X1P1  =  A1P1, 
or  that        A1B1  +  Bid  +  •  •  •  +  X1P1  +  PiAx  =  0. 

But  this  follows,  in  virtue  of  Theorem  II,  from  the  fact  that 
ABC  .  .  .  XPA  is  a  closed  broken  line. 

7.  Exercises. 

1.  Plot  the  points  A  (3,4),  B  (5,7)  and  C  (7,2).  Project  the  segments  AB, 
BC  and  CA  upon  the  X-axis  and  prove  that  the  sum  of  the  projections  is 
equal  to  zero. 

2.  Proceed  in  a  similar  manner  with  the  points  P  (1,—  3),  Q  (3,2),  R  (5,4) 
and  S  (7,-4). 

3.  Determine  the  length  of  the  projections  upon  the  X-  and  F-axes  of  the 
broken  line  O  (0,0),  A  (4,0),  P  (4,4). 

4.  Prove  that  the  sum  of  the  projections  upon  the  F-axis  of  the  segments 
AB,  BC  and  CA  of  Exercise  1  is  equal  to  zero. 

5.  Show  that  the  projections  upon  each  of  the  coordinate  axes  of  the 

straight  line  PS  of  Exercise  2  is  equal  to 
the  sum  of  the  projections  of  the  segments 
of  the  broken  line  PQRS. 

6.  The  points  0(0,0),  R  (1,2)  and  Q  (-3,4) 
(see  Fig.  5)  are  the  vertices  of  a  right  tri- 
angle.    Show   that   the   projection  of   the 

i   .  \^ >  jy     hypotenuse  OQ  upon  each  of  the  coordinate 

axes  is  equal  to  the  sum  of  the  projections 
of  the  legs  OR  and  RQ. 

7.  Plot  the  points  A  (-4,3)  and  B  (-2, 
—  7).     Show  that  the  projections  upon  each 

YIG    5  of  the  axes  of  the  line  AB  is  equal  to  the 

sum  of  the  projections  of  AO  and  OB. 

8.  Plot  the  points  Q  (  —  3,  —  7)  and  R(l,— 3).  Show  that  the  projections 
upon  the  coordinate  axes  of  OQ  is  equal  to  the  sum  of  the  projections  of  OR 
and  RQ. 

9.  Proceed  as  in  S  for  the  points  Q  (4,-4)  and  R  (-2,-3). 

8.  Directed  angles.  In  Iho  preceding  paragraphs  we  have 
enlarged  the  concept  of  line  segment  by  attributing  to  it  not  only 
magnitude,  but  also  direction.  We  proceed  now  to  extend  the 
concept  of  angle  in  a  similar  way,  viz.,  by  giving  an  angle  sense 
as  well  as  magnitude.  We  distinguish  between  the  two  sides  of 
an  angle  by  calling  one  the  initial  side,  the  other  the  terminal 


Q 


6  RADIAN  MEASURE 

side.  The  angle  is  now  to  be  thought  of  as  having  been  generated 
by  rotating  a  line  from  the  position  of  the  initial  side  to  the  position 
of  the  terminal  side;  and  the  angle  is  measured  by  the  amount  of 
this  rotation.  If  the  rotation  is  clockwise,  the  resulting  angle  is 
called  negative;  if  the  rotation  is  counterclockwise,  the  angle  is 
called  positive.  The  initial  and  terminal  sides  of  an  angle,  as  well 
as  the  sense  of  rotation,  are  indicated  by  means  of  an  arrow,  as 
in  Figure  6.  0 


The  most  common  unit  of  measurement  for  angles  is  the  degree, 
which  is  the  360th  part  of  the  angle  obtained  by  one  complete 
revolution.  An  angle  of  180°  (180  degrees)  is  called  a  straight 
angle,  an  angle  of  90°  (90  degrees)  is  called  a  right  angle.  Angles 
smaller  than  a  right  angle  are  called  acute ;  angles  greater  than  a 
right  angle  are  called  obtuse.  Two  angles  whose  sum  is  equal  to 
a  straight  angle  are  called  supplementary  angles;  two  angles 
whose  sum  is  equal  to  a  right  angle  are  called  complementary 
angles. 

9.  Exercises. 

1.  Draw  angles  of  -45°,  457°,  -312°,  583°,  1080°,  -630°. 

2.  Determine  the  complements  of  the  following  angles:  25°,  78°,  —  23 J, 
154°,  217°,  -112°,  325°,  427°,  -508°.  Construct  these  angles  and  their  com- 
plements. 

3.  Determine  the  supplements  of  the  following  angles:  89°,  127°,  —212°, 
195°,  -315°,  287°,  513°.  -459°. 

4.  Draw  the  following  angles:  120°  +  70°,  65°  +  145°,  180°  -  25°, 
270°  -  105°,  180°  +  55°,  270°  +  132°. 

10.  Radian  measure.  The  degree,  in  which  the  angles  pre- 
sented so  far  have  been  measured,  is  the  unit  commonly  used  in 
practical  work.*  For  theoretical  purposes,  another  unit,  called 
the  radian  is  to  be  preferred.     A  radian  is  an  angle,  which  sub- 

*  In  France,  the  "grade,"  one  four-hundredth  part  of  a  complete  revolu- 
tion, is  frequently  used  as  a  unit  of  measurement.  It  has  the  advantage  of 
fitting  into  the  system  of  decimal  notation. 


RADIAN   MEASURE  7 

tends  on  the  circumference  of  any  circle  described  with  the  vertex 
of  the  angle  as  a  center,  an  arc  equal  to  the  radius  of  that  circle. 


Fig.  7 

Since  the  circumference  of  a  circle  of  radius  r  is  equal  to  2  irr, 
the  radius  of  any  circle  will  be  contained  2  x  times  in  its  cir- 
cumference.    Hence  the  following  relations  hold : 

360°  =  2  7T  radians; 

1°  =  ~  radians  =  0.01745  .  .  .  radians ; 
lot) 

180° 
1  radian  =  —  =  57°.29578  •  •  •  =  57°  17'  44".81  .... 

T 

By  means  of  these  relations  the  measurement  of  an  angle  can 
be  converted  from  radian  measurement  into  degree  measure- 
ment and  vice  versa.  The  radian  measurement  of  an  angle  is 
frequently  called  the  circular  measure  of  the  angle.  When  no 
unit  is  indicated,  it  is  understood  that  circular  measure  is  meant. 
Whenever  possible,  the  circular  measure  of  an  angle  is  expressed 
by  means  of  multiples  and  submultiples  of  t* 

The  radian  measurement  of  an  angle  at  the  center  of  a  circle 
bears  a  simple  relation  to  the  arc  which  this  angle  intercepts  on 
the  circumference  of  the  circle.  For  every  radian  in  the  angle, 
we  will  have  on  the  circumference  an  arc  equal  to  the  radius. 
Hence,  if  the  radian  measurement  of  the  angle  be  denoted  by  d 
and  the  radius  of  the  circle  by  r,  we  shall  have 

length  of  arc  =  0  r. 

*  A  submultiple  of  a  number  is  the  quotient  of  that  number  by  an  integer; 
e.g.,  a/5  is  a  submultiple  of  a;   7r/3  is  a  submultiple  of  w. 


EXERCISES 


11.   Exercises. 


1.  Determine  the  circular  measure  of  the  following  angles:    45°,  30°, 
-225°,  330°,  540°.  -150°,  60°,  -270°,  -810°,  210°,  480°,  -650°. 

2.  Determine  the  number  of  degrees  in  each  of  the  following  angles: 

3  7T         — 4   IT        2  7T        5   TV         — ; 


2  3     '    5        4        2 


,  1.5,  2.3. 


3.  Determine  the  circular  measures  of  the  complements  and  of  the  sup- 
plements of  the  following  angles:  2  tt/3,  -3x/4,  75°,  7  tt/6,  -112°,  -5  tt/3, 
135°,  tt/8,  2  tt/9,  -325°,  7  tt/6,  -3  »/2. 

4.  Construct  the  following  angles:  —w/2,  5  tt/6,  7  tt/3,  —  llx/4,  ir/6, 
3  x/4,  -2  tt/3,  2  tt,  - 11  tt/6,  -3  tt,  2,  -5.2. 

6.  A  wheel  makes  10  revolutions  per  second.  Determine  the  circular 
measure  of  the  angle  described  by  one  of  the  spokes  in  15  seconds. 

6.  How  large  an  arc  will  a  central  angle  of  2.5  radians  subtend  on  a 
circle  whose  radius  is  4  feet? 

7.  How  large  an  angle,  at  the  center  of  a  circle  whose  diameter  is  10 
feet,  will  subtend  an  arc  whose  length  is  equal  to  4  feet? 

8.  An  angle  of  2  radians  placed  at  the  center  of  a  circle  intercepts  an  arc 
of  4  feet.     What  is  the  radius  of  the  circle  ? 

9.  If  a  wheel  makes  50  revolutions  per  second,  how  large  an  angle  does 
its  radius  describe  in  1  minute? 

10.   A  carriage  wheel  covers  a  distance  of  1  mile  in  1000  revolutions.     How 
large  is  its  radius? 


CHAPTER   II 
THE    TRIGONOMETRIC    RATIOS.     SIMPLE    IDENTITIES 

12.  Standard  position.  For  the  purpose  of  comparing  differ- 
ent angles  we  place  them  in  such  a  way  that  their  vertices  coin- 
cide with  the  origin  of  a  system  of  coordinate  axes,  and  that  their 
initial  sides  fall  along  the  positive  half  of  the  X-axis  of  this  sys- 
tem. Angles  so  placed  are  said  to  be  in  standard  position.  Ac- 
cording as  the  terminal  side  of  the  angle  in  standard  position 
falls  in  I,  II,  III  or  IV,  the  angle  is  said  to  lie  in  the  first,  second, 
third  or  fourth  quadrant  respectively. 

13.  Exercises. 

1.  Draw  the  following  angles  in  standard  position:  7.ri">,  —  tt/3,  120°,  v/2, 
-165°,  5  tt/6,  -21.3°,  -9  v/5,  325°,  195°,  -3  t,  155°. 

2.  Draw,  in  standard  position,  the  complements  of  the  angles  given  in 
Ex.  1. 

3.  Draw,  in  standard  position,  the  supplements  of  the  angles  given  in 
Ex.   1. 


*-X 


14.  Trigonometric  ratios.  On  tho  terminal  side  of  the  angle 
6,  placed  in  standard  position,  an  arbitrary  point  P  is  taken. 
The  coordinates  of  P  and  its  distance 
OP  from  0,  called  the  radius  vector  of 
P,  are  denoted  by  x,  y  and  r  respec- 
tively; it  is  agreed  that  the  direction 
from  0  to  P  shall  always  be  the  posi- 
tive direction  along  the  terminal  side, 
so  that  r  is  always  a  positive  number. 
While  now  the  numbers  x,  ij  and  r  will 
be  different  for  different  positions  of 
the  point  P  on  the  same  terminal  side, 

it  follows  from  the  properties  of  similar  triangles  that  any  one 
of  the  six  ratios  which  subsist  among  these  three  numbers  will 
be  the  same  for  one  position  of  the  point  P  as  for  any  other  posi- 
tion and  will  depend  upon  the  position  of  the  terminal  side  only. 

9 


Fig.  8 


10  SIGNS  OF  THE  RATIOS 

These  six  ratios  are  called  the  trigonometric  ratios  of  the  angle  0, 
and  are  fundamental  in  the  whole  field  of  mathematics.  They 
are  known  by  the  following  names: 

-  Is  called  the  sine  of  the  angle  6,  written  sin  o, 

-  is  called  the  cosine  of  the  angle  9,  written  cos  9, 
r 

-  is  called  the  tangent  of  the  angle  0,  written  tan  o, 

x 

-  is  called  the  cotangent  of  the  angle  0,  written  cot  0, 
y 

-  is  called  the  secant  of  the  angle  o,  written  sec  0, 

X 

-  is  called  the  cosecant  of  the  angle  0,  written  cosec  0. 
y 

Notice  that  the  two  ratios  occurring  in  each  of  the  three  groups 
are  obtained,  one  from  the  other,  by  replacing  x  by  y  and  y  by  x, 
and  that  interchanging  x  and  y  replaces  each  ratio  by  its  co-ratio.* 

15.  Exercises. 

1.  Draw  each  of  the  following  angles  in  standard  position,  measure  the 
ooordinates  and  the  radius  vector  of  some  point  on  the  terminal  side  of  each 
and  determine  the  trigonometric  ratios:  25°,  ir/6,  2  n-/3,  145°,  7  tt/5,  220°, 
7  jr/4,  315\  11  ir/5.  425*,  21  tt/8,  700°. 

2.  Proceed  as  in  Exercise  1  with  the  following  angles:  —30°,  x/4,  —130°, 
-5tt/6,  -210°,  -4r/3,  -320°,  -5x/4,  -112°,  -19  x/6,  -670°,  -17x/4. 

16.  Signs  of  the  ratios.  Because  the  radius  vector  is  always 
positive,  and  because  the  signs  of  the  coordinates  of  a  point 
depend  only  upon  the  quadrant  in  which  the  point  lies,  the  alge- 
braic signs  of  the  trigonometric  ratios  of  an  angle  are  determined 
by  the  quadrant  in  which  the  angle  lies.  The  agreements  as  to  the 
signs  of  the  coordinates  of  a  point  (see  3)  lead  to  the  following 
table  of  signs  for  the  trigonometric  ratios  at  the  top  of  p.  11. 

We  observe,  furthermore,  that  the  numerical  values  of  the  ratios 
of  an  angle  in  any  quadrant  are  equal  to  the  ratios  of  some  angle 
in  the  first  quadrant.  A  careful  inspection  of  the  diagram  will 
enable  the  student  to  determine,  for  any  given  angle,  an  angle 
in  the  first  quadrant  whose  ratios  are  numerically  equal  to  those 

*  For  the  origin  of  the  names  of  the  trigonometric  functions  see,  for  in- 
stance, Ctijori,  "A  History  of  Mathematics,"  p.  109. 


REDUCTION   TO   FIRST  QUADRANT 


n 


I 

n 

III 

IV 

+ 

+ 

- 

+ 

- 

- 

+ 

tangent 

+ 

- 

+ 

- 

cotangent 

+ 

- 

+ 

- 

secant 

+ 

- 

+ 

cosecant 

+ 

+ 

- 

of  the  given  angle.  By  the  use  of  a  table  of  the  trigonometric 
ratios  of  acute  angles  the  ratios  of  an  arbitrary  angle  can  there- 
fore be  found. 

17.  Exercises. 

Draw  each  of  the  following  angles  in  standard  position;  determine  the 
trigonometric  ratios  by  measurement,  as  in  14.  Determine  an  angle  in  the 
first  quadrant  whose  trigonometric  ratios  are  numerically  equal  to  those  of 
the  given  angle,  and  verify  the  results  of  the  graphical  determination  by 
means  of  a  table  of  the  trigonometric  ratios  of  acute  angles: 

1.  130°.  2.  215°.  3.  IItt/6.  4.  -493°.  5.  1300°.  6.  -9  x/5.  7.  4  tt/5. 
8.    -1105°.     9.    600°.     10.    llx/8.     11.    310°.     12.    17  x/6. 

18.  Reduction  to  first  quadrant.  We  turn  now  to  the  problem 
of  determining  for  every  arbitrary  angle  9  an  angle  9'  in  I,  such 
that  the  ratios  of  9'  shall  be  numerically  equal  to  the  ratios  of  9. 

Consider,  as  an  example,  the  angle  9  in  III,  in  Figure  9a.     If 


i 

Y 

p" 

q    _e 

\y" 

i    , 

x    y 

O      X" 

Q"   ' 

y 

/r 

i 

3 

Fig.  9a 


Fig.  9b 


we  construct  in   I  an  angle  6'  equal  in  magnitude1  to  the  acute 
angle  between  the  terminal  side  of  9  and  the  X-axis,  and  take  a 


12  EXERCISES 

point  P'  on  the  terminal  side  of  0',  such  that  OP'  =  OP,  we  find 
by  considering  the  two  triangles  OP'Q'  and  OPQ,  that 

r'  =  r,   —y'  =  y,  and  —x'  =  x.         Why? 

It  follows  from  this  that  the  ratios  of  6  are  numerically  equal 
to  those  of  6',  the  acute  angle  made  by  the  terminal  side  of  6 
with  the  X-axis. 

If  we  construct  in  I  an  angle  6"  equal  to  the  acute  angle  be- 
tween the  terminal  side  of  0  and  the  F-axis,  and  take  on  the 
terminal  side  of  6"  a  point  P"  such  that  OP"  =  OP  (see  Fig.  9b), 
we  find  upon  considering  the  triangles  OP"Q"  and  OPQ,  that 

r  =  r",     x  =  —  y",     and  y  =  —  x".  Why? 

It  follows  from  this  that  the  ratios  of  8  are  numerically  equal 
to  the  co-ratios  of  the  angle  6",  the  acute  angle  made  by  the 
terminal  side  of  6  with  the  F-axis. 

These  facts  are  summarized  in  the  following  theorem: 

Theorem  I.  The  numerical  values  of  the  trigonometric  ratios  of 
any  angle  6  are  equal  to  the  ratios  of  the  positive  acute  angle  6'  which 
the  terminal  side  of  d  makes  with  the  X-axis;  they  are  also  equal  to 
the  co-ratios  of  the  positive  acute  angle  0"  which  the  terminal  side  of 
6  make>  with  the  I -axis. 

The  algebraic  signs  of  the  ratios  are  determined  by  the  quad- 
rant in  which  the  angle  6  lies  (see  16). 

19.   Exercises. 

1.  Apply  the  theorem  of  18  to  angles  in  I  and  show  that  the  trigonometric 
ratios  of  a  positive  acute  angle  are  equal  to  the  co-ratios  of  its  complement. 

2.  Apply  the  theorem  of  18  to  angles  in  II  and  show  that  the  trigonometric 
ratios  of  an  angle  that  is  less  than  180°,  are  numerically  equal  to  those  of  its 
supplement. 

3.  Apply  the  theorem  of  18  to  angles  in  IV  and  show  that  the  trigonometric 
ratios  of  a  negative  acute  angle  are  numerically  equal  to  those  of  the  corre- 
sponding positive  angle. 

4.  Determine  all  the  trigonometric  ratios  for  the  following  angles: 


(a)  237°,                            (b)  3tt/5, 

(C)      11    TT     8, 

(d)  338°,                             (r)    1(33°, 

(/)  17  x/9. 

Also  for  the  following  angles: 

i)    -248°,   (b)    -512°,   (c)    ^,  (<D 

-11   TT 

~~6~      ' 

(*>x  (/)nr 

RATIOS  OF  30°,  60°,  45= 


13 


>A' 


20.  Ratios  of  acute  angles.  For  acute  angles,  the  definitions 
of  14  may  be  put  in  a  special  form  equivalent  to  that  of  14,  but 
frequently  better  adapted  to  the  use  that  is  to  be  made  of  them. 
The  abscissa,  ordinate  and  radius  vector  of  any  point  P  on  the 
terminal  side  of  the  angle  9  are,  in  this 
case,  the  sides  and  hypotenuse  of  a 
right  triangle  OPQ,  in  which  the  given 
angle  6  actually  occurs  as  an  acute  angle. 
The  scaffolding,  constituted  by  the 
coordinate  axes,  may  now  be  removed, 
and  the  ratios  of  the  angle  6  may  be 
defined  with  reference  to  this  triangle 
OPQ  instead  of  with  reference  to  the 
coordinate  axes,  by  putting 

"side  opposite  the  angle"  in  place  of  y, 
"side  adjacent  to  the  angle"  in  place  of  x, 
"hypotenuse"  in  place  of  r. 

In  this  way  we  obtain  the  following  form  for  the  definition  of 
the  trigonometric  ratios  of  acute  angles : 


sine  9  = 


tangent  9  = 


secant  e 


21. 


side  opposite  0 
hypotenuse 

side  opposite  e 
side  adjacent  to 

hypotenuse 
side  adjacent  to  9 


cosine  9  = 


>      cotangent  9  = 


cosecant  9 


side  adjacent  to 
hypotenuse 

side  adjacent  to 
side  opposite  9 

hypotenuse 
side  opposite  9 


Ratios  of  30°,  60°,  45°.     This  special  form  of  the  definitions 
Y  enables  us  to    find   simple    numerical 

values  for  the  ratios  of  certain  angles: 
(a)  To  determine   the  ratios  of  an 
angle  of  30°,  or  of  an  angle  of  (30°,  a 
~^*x   diagram  like  the  one  in  Figure   11  is 
constructed,  in  which  QP'  =  PQ,  and 
ZOQP  =   Z.OQP'  =  90°.      It    follows 
that  ZPOP'=  Z()!>P'=  Z  OPT  =  60°, 
so  that  A  OPP'  is  equilateral.     Choos- 
ing PQ  as  the  unit  of  measurement,  we  find  OP  =  2.  PQ  =  1,  OQ  = 
V3.     The  definitions  of  20  lead  then  to  the  following  results: 


14 


EXERCISES 


Theorem  II.    The  trigonometric  ratios  of  angles  of  30°  and  60°  are 
given  by  the  following  formulae: 

sin  30°=  1/2;     cos  30°  =  V3/2;     tan  30°  =  1/Vjj  =  V3/3     cot  30°  =  V3; 

sec  30°  =  2/ V3  =[2  V3/3;     cosec  30°  =  2; 
Sin  60°  =  V3/2;    cos  60°  =  1/2;      tan  60°  =  V3 ;      cot  60°  =  I/V3  =  V3/3; 

sec  60°  =  2;  cosec  60°  =  2/ V3  =  2  V3/3. 
p 


>x 


Fig.  12 

(6)  For  the  ratios  of  45°,  a  diagram  like  the  one  in  Fig.  12  is 
used,  in  which  OQ  is  taken  as  the  unit  of  measurement.  The 
results  are  summarized  in 

Theorem  III.    The  trigonometric  ratios  of  an  angle  of  45°  are  given 
by  the  following  formulae: 

sin  45°=  1/V2  =  V2/2;    cos  45°=  1/V2  =  V2  2;    tan  45°=  cot  45°=  1; 

sec  45°  =  cosec  45°  =  V2. 

22.    Exercises. 

1.  Determine  the  trigonometric  ratios  of  the  following  angles,  without  the 
use  of  the  tables: 

(a)   120°,  135°  and  150°.      (b)  210°,  225°  and  240°,      (c)  300°,  315°  and  330° 

2.  Determine  the  value  of  each  of  the  following  expressions,  without  the 
use  of  tables: 

(a)  sin  30°  cos  60°  +  cos  30°  sin  60°; 

(b)  cos  120°  cos  30°  -  sin  120°  sin  30°; 
C  )     tan  120°  +  tan  00°  . 

{C)    1  -  tan  120°  tarTGO0' 

(d)  cos  150°  sin  120°  -  cos  210°  sin  150°; 

3.  Similarly  for: 

(a)  sin  315°  cos2  45°  +  tan2  30°  sec  135°; 
(6)   sin2  240°  cot  225°  -  cos2  330°  tan  315°;* 

(c)  cosec2  300°  sin  60°  tan  150°  +  sec2  210°  cot  240°  cos2  30°; 

(d)  cof  150°  cosec  240°  -  tan  330°  sec3  150°. 


*  The  notation  sin2  0  means  (sin  0)2. 
powers  of  the  trigonometric  ratios. 


Similar  notations  are  used  for  other 


RATIOS  OF  90°,  180°,  270°,  360°  15 

23.  Ratios  of  90°,  180°,  270°,  360°.  For  the  ratios  of  an  angle 
whose  terminal  side  coincides  with  one  of  the  coordinate  axes,  a 
special  consideration  is  necessary,  because  application  of  the 
definitions  of  14  would  require  division  by  0,  an  operation  which 
is  impossible.  To  divide  a  number  a  by  0  would  require  the 
determination  of  another  number  b  such  that  b  multiplied  by  0 
would  yield  a,  which  is  obviously  impossible,  unless  a  were  0. 

When  8  =  0°,  we  have  x  =  r,  y  =  0.     Hence: 

sin  0°  =  0/r  =  0,    cos  0°  =  r/r  =  1,    tan  0°  =  0/r  =  0, 
cot  0°  =  r/0  =  ?,     sec  0°  =  r/r  =  1,    cosec  0°  =  r/0  =  ?. 

For  an  angle  of  90°,  x  =  0,  and  y  =  r,  so  that  we  now  find: 

sin  90°  =  1,    cos  90°  =  0,     tan  90°  -  r/0  =  ?, 
cot  90°  =  0,    sec  90°  -  r/0  -  ?,    cosec  90°  =  1. 

Similar  results  are  obtained  for  angles  of  180°  and  270°;  but 
it  should  be  noticed,  that  for  an  angle  of  180°,  the  x  and  the  r 
are  equal  numerically  but  opposite  in  sign,  in  accordance  with 
the  agreements  as  to  the  signs  of  these  quantities,  so  that  we 
have  in  this  case  x  =  —  r.  For  a  similar  reason,  we  find  that 
for  an  angle  of  270°,  y  =  —r.  Consequent^  the  ratios  of  these 
angles  have  the  following  values: 

sin  180°  =  0,      cos  180°  =  -1,      tan  180°  =  0, 

cot  180°  =  -  r/0  =  ?,     sec  180°  =  -1,    cosec  180°  =  r/0  =  ?. 

sin  270°  =  -1,    cos  270°  =  0,    tan  270°  =  -r/0  =  ?, 

cot  270°  =  0,    sec  270°  =  r/0  =  ?,     cosec  270°  =  -1. 

The  question  marks  put  after  cot  0°,  cosec  0°;  tan  90°,  sec  90°, 
cot  180°,  cosec  180°;  and  fan  270°,  sec  270°  are  intended  to  indi- 
cate that  these  ratios  cannot  be  determined  because  a  division 
by  0  would  be  involved  therein. 

These  facts  are  expressed  in  the  following  theorem: 

Theorem  IV.  The  cotangent  and  the  cosecant  of  angles  of  0°  and 
180°  do  not  exist;  the  secant  and  the  tangent  of  angles  of  90°  and 
270°  do  not  exist.  The  remaining  ratios  of  these  angles  are  given 
by  the  following  formulae: 


sin  0°  =  tan  0°  =  0, 

cos  0°  =  sec  0°  =  1 : 

sin  90°  =  cosee  90°  =  1, 

cos  9(T  =  cot  90°  =  0; 

sin  180°  =  tan  180°  =  0, 

cos  180°  =  sec  180°  =  -1; 

sin  '170°  =  cosec  270°  = 

-1, 

cos  170°  =  cot  270°  =  0. 

16 


TRIGONOMETRIC   FUNCTIONS 


24.  Trigonometric  functions.  When  an  angle  is  changing,  as, 
e.g.,  the  angle  described  by  the  spoke  of  a  revolving  wheel  or  the 
angle  described  by  a  swinging  pendulum,  there  is  in  general  a 
value  of  each  of  the  trigonometric  ratios  corresponding  to  every 
value  of  the  variable  angle.  Whenever  such  a  correspondence 
exists  between  two  varying  magnitudes,  whereby  to  each  value 
of  one  there  corresponds  a  value  of  the  other,  it  is  said  that  one  of 
the  variables  is  a  function  of  the  other.  In  the  present  case,  the 
trigonometric  ratios  are  functions  of  the  angle;  they  are  called 
trigonometric  functions. 

If  we  consider  a  variable  angle  9,  which  while  steadily  remain- 
ing in  I,  increases  towards  r/2,  the  variation  of  the  tangent  of  9 
may  best  be  studied  by  keeping  either  x  or  y  fixed  as  the  angle 
varies,  as  in  Figures  13a  and  13b  respectively.     In  either  case,  the 


*~x 


Fig.  13a 


Fig.  13b 


ratio  y/x  remains  positive;   its  numerical  value  increases  indefi 
nitely.     These  facts  are  expressed  by  the  following  statement: 
"  As  9  tend*  towards   90°,   being   always  less  than  90°,   tan  9 

increases  indefinitely,  and  by  the  formula: 

lim  tan  9  =  +  oo ." 
e— »9o°  - 

In  a  similar  manner  it  is  seen  that  if  an  angle  tends  towards 
90°,  while  steadily  remaining  in  II,  the  ratio  y/x  is  negative 
throughout,  while  the  numerical  value  increases  indefinitely;  i.e., 

"  As  9  tends  towards  90°,  being  always  greater  than  90°,  tan  9 
decreases  indefinitely,  expressed  by  the  formula: 

lim  tan  9  =  —  oo." 

e  —>  no0  + 

The  fact    that    lim    tan  9  =  +oo,   and   that   lim    tan  9  =  *=  oc 
fl—von"  -  e— ►oo0  4- 

corroborates  our  former  conclusion  that  for  an  angle  of  90°  the 

tangent  does  not  exist . 


PERIODICITY 


17 


25.   Exercises. 

1.  Interpret  and  prove  the  following  statements: 
(a)    lim  sec  0  =  -fao. 

0  — >90°- 

(c)     lim  cot  0  =  -f-ao. 

0— >180°  + 

(e)    lim  cosec  0  =  +oo. 

e  — > 180°  - 

2.  Also: 
(a)    lim  tan  0  =  +oo. 

0  — > 270°  - 

(c)     lim  sec  0  =  -f-  go. 
0  — >  270°  + 

(e)     lim  cot  0  =  +ao. 
0  — > 360°  + 


(</)     lim  cosec  0 
0  _^  360°  + 


+  *. 


(6) 

0- 

lim  sec  0  =  —  ao. 
->90°  + 

(d) 
0- 

lim  cot  0  =  —oo. 
->  180°  - 

0 

lim  cosec  0  =  —  or 

->  180°  -f 

(6) 
0 

lim  tan  0  =  —  oc. 

— ►  270°  + 

(d) 
0 

lim  sec  0  =  —00. 

-> 270°  - 

0 

lim  cot  0  =  —  x. 
->  360°  - 

(h) 

e 

lim  cosec  0  =  —  a 
—>  360°  - 

3.    Determine  the  value  of  the  following  expressions: 
(a)  sin2  jt/2  cos2  ir/6  —  tan2  3  7r/4  sec2  11  x/3. 
(6)  cos2  7r/2  sin2  3  w/2  +  cot2  11  tt/4  cosec2  11  tt/6. 

(c)  sin  3  w  cot  3  tt/2  —  tan  5  tt/6  cos  3  ir. 

(d)  sec  3  ir  sec2  9  7r/4  +  cosec  3  w/2  cosec2  17  ir/6. 

26.  Periodicity.  In  the  preceding  paragraph  we  have  dis- 
cussed the  properties  of  the  trigonometric  functions  of  the  vari- 
able angle  6  for  values  of  d  which  increase  or  decrease  towards 
0°,  90°,  180°  and  270°.  We  consider  next  some  other  properties 
of  the  trigonometric  functions,  the  first  to  be  considered  being 
the  property  of  periodicity.  A  function  is  said  to  be  periodic  if 
there  exists  a  constant  number  a,  such  that  the  function  assumes 
the  same  value  for  6  +  a  as  for  0,  no  matter  what  value  d  may  have. 
The  number  a  is  called  the  period  of  the  function.  Examples  of 
periodic  functions  are  furnished  by  many  natural  phenomena, 
such  as  the  swinging  pendulum,  the  motion  of  the  tides,  the 
movement  of  a  vibrating  string,  the  wave  motion  of  sound,  light 
and  electricity,  etc.  The  fact  that  the  trigonometric  functions 
possess  this  property  of  periodicity  is  one  of  the  reasons  for  their 
great  importance  in  the  study  of  natural  phenomena. 

If  two  angles  which  differ  by  an  integral  multiple  of  360°  (2  w) 
are  placed  in  standard  position,  they  have  the  same  terminal 
side;   hence  their  trigonometric  ratios  are  equal;  e.g., 

sin  (0  +  m  ■  360°)  =  sin  (6  +  m  •  2tt  )  =  sjn  6, 


18       RELATIONS   BETWEEN   TRIGONOMETRIC  FUNCTIONS 

for  every  value  of  6,  and  for  every  positive  or  negative  integral 
value  of  m.     We  have  therefore  the  following  theorem: 

Theorem  V.  The  trigonometric  functions  are  periodic;  they  have 
a  period  of  2  w. 

In  Chapter  V  we  shall  see  how  the  trigonometric  functions,  in 
particular  the  sine  and  the  cosine,  may  be  used  for  the  repre- 
sentation of  more  general  periodic  functions. 

27.  Relations  between  the  trigonometric  functions.  From  the 
definitions  of  the  trigonometric  ratios  in  14,  there  follow  imme- 
diate^ the  following  theorems: 

Theorem  VI.  The  sine  and  the  cosecant,  the  cosine  and  the  secant, 
the  tangent  and  the  cotangent  of  any  angle  are  reciprocals  of  each 
other;  i.e.: 

sin  e  •  cosec  0  =  1;    cos  0  •  sec  0  =  1;    tan  0  •  cot  0  =  1. 

Theorem  VII.  The  tangent  of  any  angle  is  equal  to  the  quotient 
of  its  sine  by  its  cosine;  the  cotangent  of  any  angJa  is  equal  to  the 
quotient  of  its  cosine  by  its  sine;    i.e.: 

,      „     sin  0         ,  „     cos  e 

tan  0  = -;    cot  0  =  —. — -  - 

cos0  sin0 

Since  (see  Fig.  8),  the  abscissa,  ordinate  and  radius  vector  of 
any  point  P  are  respectively  the  legs  and  the  hypotenuse  of  a 
right  triangle,  we  have  for  any  angle  6: 

x1  +  y2  —  r2. 

Dividing  both  sides  of  this  equality  in  succession  by  r2,  by  x2 
and  by  ?y2,  and  making  use  of  the  definitions  given  in  14,  we 
obtain : 

Theorem  VIII.  The  sum  of  the  squares  of  the  sine  and  the  cosine  of 
any  angle  is  equal  to  unity;  the  square  of  the  secant  of  any  angle 
diminished  by  the  square  of  the  tangent,  and  the  square  of  the  cose- 
cant of  any  angle  diminished  by  the  square  of  the  cotangent  are 
each  equal  to  unity;  i.e.: 

sin2  0  +  cos2  0  =  1;     sec2  0  —  tan2  0  =  1;    cosec2  0  -  cot2  6  =  1. 

Theorems  VI,  VII  and  VIII  may  be  used  to  determine  all  the 
trigonometric  ratios  of  an  angle  as  soon  as  one  of  them  is  known. 
By  means  of  these  theorems  a  great  many  other  relations  may  be 
proved  to  hold  between  the  trigonometric  functions. 


EXERCISES  19 


28.   Exercises. 


1.  Given  sin  0  =  —  §.     Determine  the  remaining  ratios  of  0. 

Since  sin  0  is  negative,  0  may  lie  in  III  or  in  IV.  In  either  case,  we  can 
draw  the  terminal  side  of  0,  by  determining  a  point  for  which  y  =  —  2  and 
r  =  3.  We  find  then  x  —  —  VB  or  x  =  Vo  according  as  0  is  in  III  or  IV. 
Knowing  x,  y,  and  r,  we  can  obtain  the  trigonometric  ratios  of  0. 

Proceeding  without  direct  application  of  the  definitions,  we  can  use  the 
formula  sin2  0  +  cos2  9  =  1  to  find  cos  9;  after  that  the  other  ratios  can  all 
be  determined  by  means  of  Theorems  VI  and  VII  of  27. 

2.  Determine  all  the  ratios  of  9,  when  it  is  known  that  tan  0  =  4. 

3.  Similarly,  when  cos  9  =  |  and  9  lies  in  IV. 

4.  Also  when  cot  9  =  —  7  and  9  lies  in  II. 

Prove  that  the  following  identities  result  from  the  formulae  of  27: 

5.  sin  9  =  tan  9  cos  0.  6.   cot  9  =  cos  9  cosec  0. 

7.   tan2  9  +  sin2  9  +  cos2  0  =  sec2  9.      8.    cosec  9  -  sin  9  =  cot  9  cos  0. 
9.   sin2  0  -  sin4  0  =  sin2  0  cos2  0.        10.    sec2  0  +  cot2  0  =  cosec'  0  +  tan2  9. 
11.   sin2  (9  —  cos2  0  =  sin4  0  —  cos4  9. 
tan2  (9  sec2  9-1 


12.  sin20  = 

13.  cos2  9  = 


1  +  tan2  0  sec2  0 

cot2  0  cosec2  0  —  1 


1  +  cot2  0  cosec2  0 

14.  tan  0  +  cot  0  =  sec  0  cosec  0. 

15.  sec2  0  +  cosec2  9  =  sec2  0  cosec2  0. 

16.  cosec2  0  sec2  0  =  cot2  0  sec2  0  +  cosec2  0  tan2  9. 


17. 


1  +  sin  0  cos0  4/1+ sin  0 


v/f 


cos  0  1  —  sin  0        »    1  —  sin  0 

cos  0 


18.   sec  0  -  tan  0  = 


1  +  sin  0 
sin  0 


19.  cosec  0  —  cot  9  =  , 

1  +  cos  0 

20.  2  (cos6  0  +  sin6  0)  -  3  (cos4  9  +  sin4  0)  =  -  1. 

tan  a  +  tan  8        ,  .       a 

21.  - — ■ —  =  tan  a.  •  tan  0. 

cot  a  +  cot  8 


=  \/| 


99     i/  *        v,)S0_       sin  0        _  1   —  cos  0 

'  +  cos  0  ~  1  +  cos  0  ~       sin0 

„_        cot  0      ,      . 

23. +  sin  0  tan  0  =  sec  0. 

cosec  0 

24.  sin  0  +  cot  0  cos  0  =  cosec  0. 

25.  sin  0  (cot  0  +  tan  0)   =  sec  0. 

26.  cos  0  (cot  0  +  tan  0)   =  cosec  0. 

27.  sec  0  -  cos  0  =  tan  0  sin  0. 


CHAPTER  III 

LOGARITHMS 

29.  Theory  of  Exponents.  In  order  to  facilitate  the  calcula- 
tions involved  in  the  application  of  the  trigonometric  ratios  to 
the  solution  of  problems,  we  take  up  the  study  of  logarithms.  We 
recall  first  the  following  parts  of  the  theory  of  exponents. 

(a)  A  continued  product,  all  of  whose  factors  are  equal  to  the 
same  number,  a,  is  called  a  power  of  that  number.  The  number 
a  is  called  the  base  of  the  power;  the  number  of  factors  in  the 
product  is  called  the  exponent  of  the  power.  Powers  are  classified 
according  to  their  exponents;  they  are  represented  in  abbreviated 
form  by  the  base  and  the  exponent. 

Thus  a  •  a  •  a  •  a  •  a,  the  5th  power  of  a,  is  represented  by  a5; 
a  being  the  base  of  the  power  and  5  the  exponent. 

(b)  Besides  powers  whose  exponents  are  positive  integral  num- 
bers, defined  in  (a),  we  consider  powers  whose  exponents  are  zero, 
negative  or  fractional.     Such  powers  are  defined  as  follows: 

The  zero-th  power  of  any  number  is  equal  to  1;  e.g.,  a0  =  1. 
A  power  of  a  whose  exponent  is  negative  is  the  reciprocal  of  the 
power  of  a,  whose  exponent  is  the  corresponding  positive  number: 

«-  =  \- 

a 

A  power  of  a  positive  number,  a,  whose  exponent  is  a  unit  frac- 
tion is  the  real  -positive  root  of  a  whose  index  is  equal  to  the  de- 
nominator of  the  fraction:  «1/a  =  Vo. 

Thus,  e.g.,  2~->  =  1/32;  36*  =  6;  7°  =  1. 

(c)  From  these  definitions  are  derived  the  following  funda- 
mental theorems,  usually  referred  to  as  the  "Laws  of  Exponents": 

I.  The  product  of  two  powers  of  a  number  a  is  equal  to  that 
power  of  a,  whose  exponent  is  equal  to  the  sum  of  the  exponents 
of  the  factors:   ap  X  a"  =  ap+Q. 

II.  The  quotient  of  two  powers  of  a  number  a  is  equal  to  that 
power  of  a,  whose  exponent  is  equal  to  the  exponent  of  the  dividend 
diminished  by  the  exponent  of  the  divisor:   a?  -f-  aQ  —  ap~a. 

20 


DEFINITION   OF  THE  LOGARITHM  21 

III.  A  power  of  a  power  of  a  number  a  is  equal  to  that  power 
of  a  whose  exponent  is  equal  to  the  product  of  the  exponents  of 
the  given  powers:    (ap)Q  —  ava. 

Thus,  e.g.:       V2  X  \  =  2*  X  2"2  =  2r§  =  V^; 
{/&  =  (34)s  =  3*; 
25  -T-  j$z  =  52  -z-  o-3  =  55  =  3,125. 

30.  The  use  of  exponents  in  calculation.  The  fundamental 
idea  underlying  the  use  of  logarithms  in  calculations  is  that 
the  laws  of  exponents  may  be  utilized  for  the  purpose  of  mul- 
tiplication and  division  of  numbers,  for  raising  numbers  to  a 
power,  or  for  extracting  roots.  If,  e.g.,  we  had  a  list  of  the  suc- 
cessive positive  integral  powers  of  2  from  21  to  2150,  the  product 
and  quotient  of  two  of  them,  their  powers  and  roots,  could, 
within  certain  limits,  be  found  by  addition,  subtraction,  multi- 
plication and  division.     For  example,  we  would  have: 

249  1 

OS   v    917   —    98+17   _   025.     _    _   049-63   _   0-14   _  . 

Z. A    6        -   £.  -   £.      ,     ^63    —   *  ~   &  —     214' 

(217)6  =  217x6  =  2102;  v7^44  =  (2114)  A  =  212. 

If,  therefore,  we  could  write  every  number  as  a  power  of  some 
fixed  number,  all  multiplications  and  divisions  could  be  reduced 
essentially  to  additions  and  subtractions,  while  powers  and  roots 
of  numbers  could  be  found  by  simple  multiplications  and  divisions. 

31.  Definition  of  the  logarithm.  Accordingly,  we  lay  down 
the  following  definition : 

Definition  I.  The  logarithm  of  any  number  p  with  respect  to 
the  base  «  is  the  exponent  of  that  power  of  «  which  is  equal  to  p. 

This  logarithm  is  designated  by  the  symbol  loga  p.  We  shall  con- 
sider only  logarithms  of  positive  numbers  with  respect  to  bases 
which  are  positive.* 

As  a  consequence  of  this  definition,  we  can  say: 

logj  16  =  4,  because  24  =  16;  log:f  \  =  —2,  because  3~2  =  J; 

log,  Vo  =  ',,  because  5-  =  V5;  log7  1  =  0,  because  7C  =  1. 

*  It  does  not  fall  within  the  scope  of  this  book  to  consider  whether  for  any 
number  p  and  any  base  a,  there  always  exists  an  unique  logarithm  of  p  with 
respect  to  the  base  a.  We  take  for  granted  that  if  a  and  p  are  positive,  there 
always  exists  uniquely  a  real  number  loga  p.  The  further  study  of  this  ques- 
tion belongs  to  the  Theory  of  Functions. 


22  FUNDAMENTAL  THEOREMS  ON   LOGARITHMS 

32.  Exercises. 

1.  Determine  log3  27,  log3  2V,  log3  a/3,  logs  1,  log3  y/ll,  logs  3. 

2.  Determine  log10  10,  log10  1000,  log10  .01,  logio  .0001,  logio  1. 

3.  Determine  log2  8,  log2  \/2,  log2  ye,  log2 1/V2,  log2  4/\/8. 

4.  Determine  logs  4,  log4  8,  log9  27,  log27  J,  logie  h 

5.  It  is  known  that  logio  17  =  1.23045,  logio  29.5  =  1.46982,  logio  83  =  1.91908. 
Determine  101-46*2,  101-23045,  101-91908,  103-23045,  10-46982 

6.  Determine  10  logl°  7,  5  Iogi  7,  2  log^ 7,  a  loga  7. 

33.  The  fundamental  theorems  on  logarithms.  From  the 
"Laws  of  Exponents,"  quoted  in  29  (c),  we  derive  now,  by  the  aid 
of  the  definition  of  31,  the  following  fundamental  theorems  on 
logarithms,  in  which  we  prove  what  was  merely  suggested  in  the 
last  paragraph  of  30. 

Theorem  I.  The  logarithm  of  the  product  of  two  numbers  with 
respect  to  the  base  a  is  equal  to  the  sum  of  the  logarithms  of  the 
factors: 

lOga  (pq)  -  lOga  p  +  lOga  q. 

Proof.  The  theorem  evidently  says  nothing  else  than  that  the 
exponent  of  that  power  of  a  which  equals  pq  is  equal  to  the  sum 
of  the  exponents  of  the  powers  of  a  which  are  equal  to  p  and  q. 
For,  let 

loga  p  =  x,  and  loga  q  =  y; 

i.e.,  let  ax  =  p,  and  av  =  q. 

Then     axJrV  =  pq;     i.e.,    logs  pq  =  x  +  y  =  loga  p  +  loga  q. 

Theorem  II.  The  logarithm  of  the  quotient  of  two  numbers  with 
respect  to  the  base  a  is  equal  to  the  logarithm  of  the  dividend  dimin- 
ished by  the  logarithm  of  the  divisor: 

lOga  -  =  lOgapJ-  lOga  q. 

Theorem  III.  The  logarithm  of  a  power  of  a  number  with  respect 
to  the  base  a  is  equal  to  the  exponent  of  the  power  multiplied  by  the 
logarithm  of  the  number: 

loga  pn  =  n  ■  log,,  p. 

The  proofs  of  Theorems  II  and  III  are  left  to  the  student. 
We  proceed  to  illustrate  the  use  of  these  theorems: 
(a)  To  determine  logio  14  when  it  is  known  that  logio  2  =  .30103 
and  logio  7  =  .84510. 


EXERCISES  23 

From   Theorem    I,   it   follows   that   logio  14  =  logio  2  +  logio  7 
=  1.14613. 

14  X  25 
(6)  To  determine  logio  — ^ — ,  when  logl0  2  =  .30103,  logio  3 

=  .47712,  logio  5  =  .69897  and  logio  7  =  .84510. 

By  the  use  of  Theorems  I  and  II,  we  find: 

logio  — 27 —  =  lo&10  14  "+*  lo&10  25  ~~  logw  Q7. 

But,  from  Theorem  III  it  follows  that  logio  25  =  logio  52  =  2 
logio  5  =  1.39794,  and  that  logio  27  =  logio  33  =  3  logio  3  =  1.43136. 

.-.    logio  H  *  2°  =  logio  14  +  2  logio  5-3  logio  3  =  1.11271. 


3/25  x  32 
(c)  To  determine  log  y  — ^ — -,  when  logio  2  =  .30103,  logio  3 

=  .47712  and  logio  5  -  .69897. 

By  the  use  of  Theorems  I,  II  and  III,  we  find: 


logio  y — ^ =  3  [2  log10  5  +  5  logio  2-4  logio  3] 

=  I  [1.39794  +  1.50515  -  1.90848]  =  |  .99461 
=  .33154. 

34.   Exercises. 

Given,  that  log10  2  =  .30103,  logl0  3  =  .47712,  and  logio  7  =  .84510,  determine 

t5/27r>< 


1. 

logio  5. 

2. 

logio  21. 

3. 

logm  35. 

4. 

log!-,  70. 

5. 

1         40 

logio  gj 

6. 

logio  \/32. 

7. 

,         125  X  32 
logio 27 

8. 

,         t  /28  X  25 
logio  V       81 

10.    logio 


-   X49 
9.   lognj  V  — ^ 

81  X  64 

25  X  </2 

...     ,        \/8  x  vT) 

11.    logio rji 

V  < 


10     1         V  32  X  75 
12.    logio  — , 

V245  X  54 


24 


COMMON  LOGARITHMS 


35.  Common  logarithms.  It  will  now  be  recognized  that  the 
plan  suggested  in  30  can  actually  be  carried  out,  provided  we  can 
"write  every  number  as  a  power  of  some  fixed  number,"  i.e., 
provided  we  can  find  the  logarithm  of  every  number  with  respect 
to  some  fixed  base.  The  Differential  Calculus  furnishes  methods 
for  solving  this  problem  and  for  constructing  a  "table  of 
logarithms."  Por  the  purpose  of  computation,  logarithms  with 
respect  to  the  base  10,  called  common  logarithms  are  most  useful. 
Without  considering  here  the  problem  of  constructing  a  table  of 
common  logarithms,  we  shall  see  how  a  comparatively  small  table 
of  logarithms  can  be  made  to  serve  our  purpose.* 

From  now  on,  the  symbol  "log"  will  be  used  to  designate  logw, 
i.e.,  common  logarithm. 

We  know  then  that  log  1000  =  3,  log  100  =  2,  log  10  =  1, 
log  1  =  0,  log  .1  =  -  1,  log  .01  =  -  2,  log  .001  =  -  3,  etc.;  i.e., 
if  we  arrange  numbers  in  the  geometrical  progression  of  scale  A, 
their  common  logarithms  will  form  the  arithmetical  progression  of 
scale  B. 


Scale  A 

JV 

.001 

.01 

.1 

1 

10 

100 

1000 

Scale  B 

log  N 

-3 

-2 

-1 

0 

1 

2 

3 

We  assume  now,  without  proof,  that  as  the  number  N  increases, 
its  logarithm  will  also  increase.  It  will  then  follow  that  if  a  num- 
ber N  lies  between  two  successive  terms  of  scale  A,  log  N  will  lie 
between  the  two  corresponding  integers  of  scale  B.  Hence  if  we 
have  determined  the  place  of  a  number  N  in  scale  A,  we  shall  know 
the  integral  part  of  log  N,  and  conversely,  if  we  know  the  integral 
part  of  log  N,  we  shall  know  the  position  of  N  in  scale  A. 

The  integral  part  of  the  logarithm  of  a  number  is  called  the 
characteristic;  the  fractional  part,  taken  positively,  is  called  the 
mantissa.  Here  it  is  to  be  understood  that  if  the  logarithm  of  a 
number  is  negative,  it  will  be  written  as  the  sum  of  a  negative 
integer  (the  characteristic)  and  a  positive  fraction  (the  mantissa). 
1 


For  example,  log 


V10 


log  10   2  =  —2  will  be  written  in  the 


*  Any  set  of  five-place  tables  of  logarithms  of  numbers  and  of  trigonometric 
functions  may  be  used  in  connection  with  this  text. 


COMMON  LOGARITHMS  25 

form  —  1  +  .50000,  or  9.50000  —  10;  the  characteristic  is  —1  (or 
9-10),  the  mantissa  is  .50000.* 

We  conclude  that  the  position  of  a  number  in  scale  A  and  the 
position  of  its  logarithm  in  scale  B  are  related  to  each  other  in 
the  way  expressed  by  the  following  theorem: 

Theorem  IV.  The  characteristic  of  the  common  logarithm  of  a 
number  .V  is  the  lesser  of  the  two  integers  in  scale  B,  corresponding 
to  the  two  numbers  in  scale  A  between  which  V  lies;  conversely,  a 
number  V  will  lie  between  those  two  numbers  in  scale  A,  which  cor- 
respond to  the  characteristic  of  log  N  and  the  next  greater  integer 
in  scale  B. 

This  theorem  may  be  reduced  at  once  to  the  following  working 
rule : 

The  characteristic  of  log  A  is  equal  to  the  integer  in  scale  B,  corre- 
sponding to  that  number  in  scale  A  which  is  pointed  off  like  A";  con- 
versely, JV  is  to  be  pointed  off  like  that  number  in  scale  A,  which 
corresponds  to  the  characteristic  of  log  Ar  in  scale  B. 

It  remains  to  determine  the  mantissa  of  a  logarithm;  it  is  found 
from  a  table.  It  is  important  to  observe  however  that  the 
mantissa  of  the  logarithm  is  the  same  for  all  numbers  which  differ 
only  in  the  position  of  the  decimal  point.  Let  us  compare,  for 
instance,  log  475.3  and  log  .04753.  Since  475.3  =  104  X  .04753, 
it  follows  that  log  475.3  =  4  +  log  .04753.  But  the  addition  of 
4  to  log  .04753  will  simply  increase  its  characteristic  by  4,  with- 
out affecting  the  mantissa.  In  general,  if  A  and  B  differ  only  in 
the  position  of  the  decimal  point,  the  greater  one,  say  A,  can  be 
obtained  by  multiplying  the  smaller  one,  B,  by  a  power  of  10,  i.e. 
.1  =  10*  X  B,  where  k  is  a  positive  integer.  Therefore  log 
A  =  k  +  log  B,  so  that  log  A  is  obtained  from  log  B  by  adding 
to  it  the  integer  k.  This  however  will  merely  increase  the  charac- 
teristic by  k  and  will  not  alter  the  mantissa.  This  fact  finds  expres- 
sion in  the  following  theorem: 

Theorem  V.  The  mantissa  of  the  common  logarithm  of  a  num- 
ber is  determined  by  its  sequence  of  digits  only;  conversely,  the 
sequence  of  digits  of  a  number  is  determined  by  the  mantissa  of  its 
logarithm. 

':  When  a  five-place  table  is  used,  all  mantissas  are  written  out  to  five 
places  of  decimals. 


26  USE  OF  A  TABLE  OF  LOGARITHMS 

36.  Use  of  a  table  of  logarithms.  Theorems  IV  and  V  enable 
us  to  find  from  a  five-place  table  of  logarithms  the  logarithm  to 
five  places  of  any  number  of  five  significant  figures;  *  and  to 
determine  five  significant  figures  of  a  number  whose  mantissa  is 
given  to  five  places.  It  only  remains  to  become  familiar  with  the 
arrangement  of  the  tables.  This  is  best  explained  by  means  of 
examples. 

Example  1.     To  find  log  47.316. 

Since  47.316  lies  between  lOjand  100,  (or,  since  47.316  is  pointed 
off  like  10)  the  characteristic  of  log  47.316  is  1.  To  determine  the 
mantissa  we  look  up  in  the  tables  the  sequence  of  digits  47316. 
The  first  three  digits  are  found  in  the  left-hand  column  of  the 
table  of  logarithms  of  numbers,  usually  headed  N,  the  fourth  digit 
is  found  in  the  line  at  the  top  or  bottom  of  the  page.  In  this  way 
we  find  that 

for  47320  the  mantissa  is  67504;  while 
for  47310  the  mantissa  is  67495. 

For  an  increase  of  10  in  the  number,  there  is  therefore  an  in- 
crease of  9  in  the  mantissa.  Now  we  approximate  to  the  value 
of  the  mantissa  for  47316  by  assuming  that  as  the  number  in- 
creases by  ten  equal  steps  from  47310  to  47320,  the  mantissa 
will  also  increase  by  ten  equal  steps  from  67495  to  67504.  We 
must  then  add  to  67495  six  tenths  of  9.  The  table  of  proportional 
parts  (usually  headed  P.P.)  shows  that  T%-  of  9  is  equal  to  5.4, 
which  we  round  off  to  5.  We  conclude  that  the  mantissa  for 
47316  is  67500  and  that  log  47.316  =  1.67500. 

Example  2.     To  find  log  .089327. 

Since  .089327  lies  between  .01  and  .1,  (i.e.  since  .089327  is  pointed 
off  like  .01)  the  characteristic  of  the  logarithm  is  —2  or  8  —  10. 

The  mantissa  for  89330  is  95100;  the  mantissa  for  89320  is 
95095.  A  difference  of  10  in  the  number  makes  a  difference  of 
5  in  the  mantissa. 

From  the  table  of  proportional  parts,  we  find  that  -^  of  5  is  3.5, 
which  we  round  off  to  4,  so  that  the  mantissa  for  89327  is  95099 
and  therefore  log  .089327  -  8.95099  -10. 

*  The  figures  of  a  number  significant  for  its  logarithm  are  the  digits  left 
after  the  ciphers  at  the  beginning  and  end  of  the  number  have  been  removed. 


USE  OF  A  TABLE  OF  LOGARITHMS  27 

Example  3.     Given  log  AT  =  2.23130.     To  determine  N. 

We  now  reverse  the  process  followed  in  the  preceding  examples 
and  begin  by  searching  among  the  mantissas  given  in  the  table 
for  23130.     We  find  that 

23147  is  the  mantissa  for  17040,  and  that 

23121  is  the  mantissa  for  17030,  so  that  a  difference  of 

26  in  the  mantissa  makes  a  difference  of  10  in  the  number. 

The  given  mantissa  23130  exceeds  23121  by  9;  from  the  table 
of  proportional  parts,  we  find  that  -^  of  26  equals  7.8,  while 
yV  of  26  equals  10.4.  Of  these,  the  former  is  closer  to  9  than 
the  latter;  hence  23130  is  the  mantissa  for  17033,  which  must  there- 
fore be  the  sequence  of  digits  for  the  required  number  N.  Since 
the  characteristic  is  2,  N  must  be  pointed  off  like  100  so  that  the 
decimal  point  must  be  placed  between  the  0  and  the  3.  We  con- 
clude that  the  required  number  N  is  170.33. 

Example  4.     Given  log  A"  =  9.07025  —  10.     To  determine  N. 

We  find  that  07041  is  the  mantissa  for  11760,  while  07004  is 
the  mantissa  for  11750,  so  that  a  difference  of  37  in  the  mantissa 
makes  a  difference  of  10  in  the  number.  Hence  a  difference  of  21 
in  the  mantissa  corresponds  to  a  difference  of  6  in  the  number; 
the  sequence  of  digits  of  N  must  therefore  be  11756.  Since  the 
characteristic  is  —1,  the  number  must  be  pointed  off  like  .1,  so 
that  we  conclude  that  N  =  .11756. 

The  table  of  the  logarithms  of  the  trigonometric  functions  gives 
directly  the  logarithms  of  the  trigonometric  ratios,  both  charac- 
teristic and  mantissa,  except  that  from  the  logarithms  of  sines 
and  cosines,  and  from  the  logarithms  of  tangents  of  angles  less 
than  45°  and  of  cotangents  of  angles  greater  than  45°,  the  term 
—  10  is  usually  omitted.  The  arrangement  of  the  table  will  be 
readily  understood  from  the  following  examples. 

Example  5.     To  determine  log  tan  37°  23'  45". 

We  find  log  tan  37°  23'  -  9.88315  -  10  and  log  tan  37°  24' 
=  9.88341  —  10,  so  that  a  difference  of  1  minute  in  the  angle 
causes  a  difference  of  26  in  the  logarithm  of  its  tangent.  Accord- 
ingly a  difference  of  45"  in  the  angle  will  cause  a  difference  of 
20  in  the  logarithm  of  the  tangent.  Hence  log  tan  37°  23'  45"  = 
9.88335  -  10. 


28  CALCULATION  BY  MEANS  OF  LOGARITHMS 

Example  6.     To  determine  log  cos  54°  29'  13". 

We  find  log  cos  54°  29'  =  9.76413  -  10,  while  log  cos  54°  30' 
=  9.76395  —  10,  so  that  as  the  angle  increases  by  1  minute,  the 
logarithm  of  its  cosine  decreases  by  18.  Consequently  for  an 
increase  of  13"  in  the  angle  there  will  be  a  decrease  of  {$  of  18, 
i.e.,  of  4  in  the  logarithm  of  the  cosine;  hence  log  cos  54°  29'  13" 
=  9.76409  -  10. 

Example  7.     Given  log  sin  6  =  9.47468  —  10.     To  determine  0 

We  find  that  log  sin  17°  21'  =  9.47452  -  10,  while  log  sin  17°  22' 
=  9.47492  —  10,  so  that  a  difference  of  40  in  the  logarithm  of  the 
sine  corresponds  to  a  difference  of  1  minute  in  the  angle.  Conse- 
quently, a  difference  of  16  in  the  logarithm  of  the  sine  corresponds 
to  a  difference  of  \%  of  a  minute,  i.e.,  of  24"  in  the  angle.  We 
conclude  that  d  =  17°  21'  24". 

37.  Exercises. 

Determine: 

1.  log  56.387.  5.  log  978.94.  9.  log  cos  47°  58'  15". 

2.  log  .084923.  6.  log  .00073299.  10.  log  cot  15°  47'  50". 

3.  log  1.0576.  7.  log  sin  27°  15'  20".  11.  log  sin  78°  29'  40". 

4.  log  .20458.  8.  log  tan  68°  37'  35".  12.  log  cos  36°  35'  45". 

Determine  the  number  N,  when : 

13.  log  JV  =  1.65783.  15.   log  N  =  .27586.         17.   log  AT=  7.80880 -10. 

14.  log  N  =  9.04987  - 10.   16.    log  N  =  .09675.  18.    log  AT  =3.97538. 

Determine  the  angle  0,  when: 

19.  log  tan  9  =  .27725.  22.  log  cot  9  =  8.83225  - 10. 

20.  log  cos  9  =  9.88247  -  10.  23.  log  cos  0  =  9.68552-10. 

21.  log  sin  9   =  9.48030  -  10.  24.  log  tan  0  =  9.96795  -10. 

38.  Calculation  by  means  of  logarithms.  We  are  now  prepared 
to  apply  the  results  of  the  preceding  articles.  In  all  calcula- 
tions, it  is  important  to  arrange  the  work  in  such  a  way  as  to 
secure  the  greatest  possible  accuracy  with  least  effort.  This  is 
accomplished  best  by  making  a  plan  of  the  entire  calculation 
before  looking  up  the  logarithms,  as  illustrated  in  the  following 
examples: 


CALCULATION   BY  MEANS  OF  LOGARITHMS  29 


Example  1.     To  determine  Ar  =  \7  -  : ^=-^ 

First  we  apply  Theorems  I,  II  and  III;  in  this  way  we  find 
that  log  N  =  f  [log  47.321  +  log  .015732  -  log  .9763]. 
Accordingly  we  make  the  following  plan  for  the  calculation: 


log  47.321  =  1 

log  .015732  =  8 -10 


A 


log  .9763  =  9 -10 

— S 

3 

\ogN  =  

N  =  


After  having  completed  the  plan,  we  turn  to  the  tables  to  deter- 
mine the  mantissas  and  to  complete  the  calculation;  this  gives 
the  following  results: 

log  47.321  =    1.67505 

log  .015732  =   8.19678  -  10 

A 


19.87183  -  20 
log  .9763  =  9.98958  -  10 


■S 


29.88225  -  30 

3 

log  Ar  =    9.96075  -  10 

Ar  =      .91358. 

In  order  to  make  the  subtraction  possible  without  introducing 
a  negative  mantissa,  we  wrote  the  characteristic  of  the  minuend 
in  the  form  19  —  20  instead  of  9  —  10:  to  make  the  division  possible 
without  introducing  a  negative  fraction,  we  wrote  the  mantissa 
of  the  dividend  in  the  form  29  -  30,  instead  of  9  -  10. 

v  7    o      t     i  +        •       ^     f356.12x  C56836)-7' 

Example  2.      lo  determine  A  =  • 

51.834  X  \  .0843 


30  EXERCISES 

We  find  that  log  N  =  5  [log  356.12  +  2 log .56836  -(log  51.834 
+  ^  log  .0843)],  so  that  the  calculation  is  carried  out  in  the  fol- 
lowing form: 

log  356.12  =    2.55159 
log  .56836  =  9.75462  -  10;    2  log  .56836  =  19.50924-20 


2.06083 
log  51.834  =  1.71461 

log'\0843  =  28.92583 -30;   |  log  .0843  =  9.64194  -  10 


1.35655 1.35655 


.70428 


log  Ar  =    3.52140 
N  =    3322.0 

7    _      _     ,  A        .      „      .075869  X  sin  47°  15'  36" 

Example  S.     I  o  determine  J\  = — ,.00  nn, 

^  tan  68  23 

log  N  =  log  .075869  +  log  sin  47°  15'  36"  -  log  tan  68°  23'. 
log  .075869  =  8.88006  -  10 

log  sin  47°  15'  36"  =  9.86596  -  10 

A 

8.74602  -  10 
log  tan  68°  23'  =     .40201 

S 


log  N  =  8.34401  -  10 
N  =     .022081 


39.   Exercises. 

Calculate: 


14.03  X  1-028 

(.005734)* 


3  A5  ^15  7    4-3857  X  .1296 

2'   V~507~'  .00097135 


4.357  X  (.08356)3  *  76.7248  X  .098376 

\/XJ57  V  57-S4y  X  -0001574 ' 

.      (.4316)*  X  52.07  _    51.86  X  sin  35°  IS' 


C083)2  sin  49°  23' 

46 1 54.7  X  -0287  10.    tan  28°  37'  15"  X  cos  56°  13' X 

V  15.8 


cosec  75°  20'  20". 


EXERCISES  31 

11.  9.63  X  sin  42°  17'  X  sin   18°  i /57.8  X  67.84.5  X  23.593 


29'  30".  '    V                  98.627 

1875  X  tan  65°  34'  10"  *  /.0024856  X  .57321  X  .009847 

sin  17°  54' 20"  17-    V~           sin  9°  15'  15" 

i /4.268  X  sin  81°  20'  t  7.08734  X  .09586  X  .06792 

'    V        tan  14°  38'        '  V                   .15647 

14     p8.643  X  cos  21°  17'~|3  .5487  X  sin  38°  27'  15" 

"    1.9.5064  X  cot  68°  23'J  "  .98346 

,  c     1.0567  X  cos  28°  43'  50"  nn    3.75  X  sin  53°  27'  30" 

!*>•    ^  „^„, •  20. 


12 


13 


2.3981  sin  145°  35'  40" 


CHAPTER  IV 
THE  SOLUTION  OF  RIGHT  TRIANGLES.    APPLICATIONS 

40.  The  right  triangle.  If  enough  elements  of  a  right  triangle 
are  given  to  determine  it  completely,  every  other  magnitude  con- 
nected with  the  triangle  can  be  determined  by  means  of  the  trigo- 
nometric ratios  of  the  angles.  A  right  triangle  is  completely 
determined  by  any  one  of  the  following  sets  of  data: 

(a)  two  legs,  (6)  one  leg  and  the  hypotenuse,  (c)  one  acute 
angle  and  a  leg,  (d)  one  acute  angle  and  the  hypotenuse. 

In  the  case  of  the  right  triangle  the  remaining  elements  can  be 
determined  in  each  case  by  the  use  of  the  special  form  of  the 
definitions  for  acute  angles,  given  in  20.  It  is  customary  to 
demote  the  lengths  of  the  sides  of  a  triangle  by  small  Roman 
letters,  the  vertices  of  the  opposite  angles  by  the  correspond- 
ing Roman  capitals,  the  vertex  of  the  right  angle  and  the  hy- 
potenuse in  a  right  triangle  usually  being  denoted  by  C  and  c, 
respectively. 

In  cases  (a)  and  (6)  the  knowledge  of  the  sides  enables  us  to 
find  one  of  the  ratios  of  either  of  the  acute  angles,  and  hence 
these  angles  themselves.  In  cases  (c)  and  (d)  we  proceed  in  the 
opposite  manner:  from  the  given  side  and  appropriate  trigono- 
metric ratios  of  the  given  angle  the  other  sides  are  found.  The 
second  acute  angle  is  the  complement  of  the  given  angle. 

From  the  definitions  given  in  20  we  obtain  immediately  the 
following  theorem: 

Theorem  I.  In  a  right  triangle,  the  following  relations  hold  be- 
tween the  sides  and  angles: 

the  side  opposite  an  acute  angle  =  hypotenuse  X  the  sine  of  the 
angle; 

the  side  adjacent  to  an  acute  angle  =  hypotenuse  X  the  cosine  of 
the  angle; 

the  side  opposite  an  acute  angle  =  the  adjacent  side  X  the  tangent 
of  the  angle. 

32 


ACCURACY  OF  THE  CALCULATION  33 

41.  Example  (see  Fig  14). 

Given.     Z.C  =  90°,  ZA  =  47°  13',  b  =  14.35. 

Required.     B,  a,  c. 

Solution.  We  find  ZB  =  90°  -  47°  13'  =  42°  47'. 

Moreover,  we  have  by  Theorem  I, 
a  =  b  tan  A ; 
also,  b  —  c  cos  A , 

whence  c  =  b/cos  A. 

The  unknown  elements  of  the  triangle  have 
now  been  expressed  in  terms  of  the  known 
elements.     It   remains  to  calculate  a  and  c.  A*~       b~u.iro 
This  may  be  done  by  the  use  of  logarithms  Fig.  14 

(see  38)  as  follows: 

log&  =  log  14.35  =    1.15685;  log  6  =  log  14.35=    1.15685 
log  tan  A  =  log  tan  47°  13'=     .03364; 

log  cos  A  =  log  cos  47°  13'  =   9.83202  - 10 
___ __^4  —S 

loga=    1.19049  logc=    1.32483 

a  =15.506  c  =  21.127 

Using  tables  giving  the  values  of  the  ratios  themselves  (the 
so-called  natural  values)  instead  of  those  of  their  logarithms,  we 
obtain : 

a  =  14.35  X  tan  47°  13'  =  14.35  X  1.0805  =  15.505. 
b  =  14.35  -f-  cos  47°  13'  =  14.35  -r-    .6792  =  21.128. 

42.  Accuracy  of  the  calculation.     Checking  the  results.     The 

values  of  the  logarithms  and  of  the  trigonometric  ratios  found  in 
a  table  are  correct  only  to  within  the  limit  of  accuracy  of  the  table, 
i.e.,  to  within  .000005  if  5-place  tables  are  used,  or  .00005  in  the 
case  of  4-place  tables. 

Therefore,  since  the  sum  and  difference  of  two  approximate 
numbers  are  more  readily  determined  and  are  frequently  more 
nearly  accurate  than  their  product  or  quotient,  it  follows  that  in 
most  cases  the  calculation  by  means  of  logarithms  is  to  be  pre- 
ferred, particularly  if  the  data  of  the  problem  are  themselves 
approximations.* 

*  A  comparison  of  the  advantages,  of  calculations  with  and  without  the  use 
of  logarithms  requires  a  much  more  detailed  discussion  than  can  here  he  de- 
voted to  it.  Most  of  the  calculations  in  this  hook  have  heen  made  hy  means 
of  logarithms,  on  account  of  the  greater  convenience  of  this  method. 


34 


ISOSCELES    TRIANGLES 


To  secure  a  higher  degree  of  certainty  as  to  the  correctness  of 
the  final  numerical  results  of  a  calculation,  these  results  should 
be  checked.  A  rough  check  can  be  obtained  by  drawing  the 
figure  to  scale  by  use  of  ruler  and  compass  and  then  measuring 
the  required  elements.  A  more  accurate,  numerical  check  in- 
volves the  testing  of  the  results,  together  with  the  data,  in  other 
relations  between  the  sides  and  angles  of  the  triangle  than  those 
used  in  the  solution  of  the  problem. 

For  the  right  triangle,  the  Pythagorean  relation  c2  =  a2  -\-  b2 
serves  the  purpose  especially  well,  since  it  involves  all  three  sides 
of  the  triangle.     If  written  in  the  form 

a2  =  c2  —  b2  =  (c  —  6)  (c  +  6)  or  b2  =  c2  —  a2  =  (c  —  a)  (c  +  a), 

it  is  well  adapted  to  logarithmic  calculation.     For  the  example  of 
the  preceding  section,  it  furnishes  the  following  check: 

c  =  21.127 
a  =  15.506 


A 


c  +  a  =  36.633  log  (c  +  a)  =  1.56388 

c-a=    5.621  log  (c  -  a)  =    .74981 

log  (c  +  a)  (c-  a)  =  2.31369 

2 

log  b  =  1.15685 

This  value  should  agree  to  within  four  units  of  the  last  decimal 
place  with  the  value  of  log  b  derived  from  the  value  of  b  as  given. 

43.  Isosceles  triangles.  An  isosceles  triangle  is  divided  into 
two  congruent  right  triangles  by  a  perpendicular  from  the  vertex 
to  the  base.  The  methods  explained  in  40,  41  and  42  suffice 
therefore  for  the  treatment  of  such  triangles. 


a 

\h 

^v"  -  -251B2 

y^fd 

^Nq    ■ 

V 

1 

Fig.   15 

Example  (see  Fig.  1 5) 

Given.      ZP  =  ZQ  = 

37 

0  28',  a  = 

.23152. 

Required.     V,  h,  b. 

EXERCISES  35 

Solution.     We   find    Z  V/2  =  90°  -  37°  28'  =  52°  32';    .'.  Z  V 
=  105°  4'. 
Furthermore 

sin  P  =  h/a,       i.e.,  /i  =  a  sin  P, 

and              cos  P  =  \  b/a,    i.e.,  ^  6  =  a  cos  P, 

whence  6  =  2  a  cos  P. 

log  a  =  9.36459  -  10  log  a  =  9.36459  -  10 

log  sin  P  =  9.78412  -  10  log  cos  P  =  9.89966  -  10 


A 


log  h  =  9.14871  -  10        log  2  =  .30103 


h  =    .14087  log  6  =  9.56528  -  10 

b  =  .36752 

Check,     /i2  =  a2  -  (6/2)2  =  (a  +  6/2)  (a  -  6/2) 
a  =  .23152 
6/2  =  .18376 


a  +  6/2  =  .41528;  log  (a  +  6/2)  =  9.61834  -  10 
a  -  6/2  =  .04776;  log  (a  -  6/2)  -  8.67906  -  10 

A 

log  [a2  -  (6/2)2]  =  8.29740  -  10 

2 

log/i  =  9.14870-  10 

44.   Exercises. 

Calculate  the  unknown  elements  of  the  triangles  ABC  in  which  Z  C  «=  90° 
and  in  which  the  following  elements  are  given;   check  the  results: 

1.  a    =  373,        6  =  52G.  7.  A  =  84°  35',  c  =  378. 

2.  a    =  .1432,     6  =  .0756.  8.  A  =  44°  35',  c  =  378. 

3.  a    =2.146,     c  =  4.292.  9.  .1   =  45°  3'.    a  =.08512. 

4.  b    =  13.071,  c  =  19.603.  10.  a   =  .06891,  c  =  .09004. 

5.  A  =  68°  25',  b  =  8732.  11.  b    =  13.683,  a  =  3.9857. 

6.  B  =  27°  13',  6  =  .06315.  12.  B  =  5°  2',      c  =  1.0059. 

Determine  the  remaining  elements  of  the  following  isosceles  triangles  (the 
notation  being  in  accordance  with  Fig.  15)  and  check  the  results: 

13.  b  =  26.804,  P  =  57°  13'.  17.  b  =  24.192,  h  =  16.387. 

14.  b  =  35.96,     V  =  128°  46'.  18.  h  =  .05831,  P  =  10°  19'. 

15.  a  =  12.05,    h    =  8.041.  19.  b  =  9.0834,  a  =  9.9457. 

16.  h  =  1.0203,  V  =  44°  52'.  20.  a  =  6.8032,  P  =  25°  27'. 

21.    a  =  16584,  V  =  90°. 


36 


PROJECTION 


45.  Projection.  The  projection  of  a  directed  segment  AB  of 
a  directed  line  I  upon  another  directed  line  m,  which  makes  with 
I  an  angle  d  can  now  be  readily  determined.  Let  the  length  of 
AB  (see  Fig,  16)  be  equal  to  a;   then,  since  the  direction  of  AB 


Fig.  16 


>■  m 


is  opposite  to  the  positive  direction  upon  I,  AB  =  —  a.  The 
projection  of  A B  upon  m  is  AiBi.  Through  A  draw  a  line  m' 
parallel  to  m,  cutting  BBX  in  C.  Then  AC  =  A,B,  (Why?), 
and  Z  CAD  =  d  (Why?).  Moreover,  cos  6  =  -cos  6'  (see  18). 
Since  now  A\BX  is  in  the  positive  direction  upon  m,  we  have 

AiBi  =  AC  =  a  cos  6'  =  —a  cos  d  =  AB  cos  6. 

This  result  finds  expression  in  the  following  theorem: 

Theorem  II.  The  projection  of  a  directed  segment  AB  of  a  directed 
line  /,  upon  a  directed  line  m,  Is  equal,  in  magnitude  and  direction, 
to  AB  cos  d,  where  9  is  the  angle  which  I  makes  with  m. 


Fin.  17 

This  theorem  is  illustrated  by  the  following  examples: 

(a)  In  Figure  17,  /  makes  with  m  an  angle  of  240°;   since  AB 

is  a  negative  segment  of  the  directed  line  I,  we  have  AB  =  —4. 

Hence,  AJi,  =  Vvo]mAB  =  -  4  •  cos  240°  =  +  2. 


APPLICATION  OF  THEOREM   II 


37 


(6)  In  Figure  18,  I  makes  with  m  an  angle  of  135°  and  AB  =  3. 

—  3\/2 

Hence  A1Bl  =  ~ProjmAB  =  3  •  cos  135°  =       n       =  -  2.1. 


Fig.  18 


46.  Application  of  Theorem  II.  We  return  now  to  Theorem  II 
of  6  and  calculate  the  projections  of  the  segments  b}T  means  of 
Theorem  II  of  the  present  chapter.  Consider,  for  example,  the 
equilateral  triangle  ABC  of   Figure   19,   of  which  the  side  AC 


Fig.  19 

makes  an  angle  of  15°  with  the  positive  X-axis. 

The  side  AB  then  makes  with  OX  an  angle  of  75°, 
the  side    BC  makes  with  OX  an  angle  of  —45°, 
the  side    CA  makes  with  OX  an  angle  of  1!).")°. 

If  the  length  of  the  side  of  the  triangle  be  denoted  by  a  and  the 
projections  of  the  vertices  upon  the  A'-axis  by  A\,  Bh  and  C\,  then 
we  find: 

ArB,  +  B1Ci  +  CiA  1  =  Proj^l  B  +  Proj^C  +  Proj^C^ 

=  a  (cos  75°  +  cos  45°  -  cos  15°) 
=  a  (.2088  +  .7071   -  .<H>r>9)  =  0. 


38 


APPLICATIONS 


47.  Exercises. 

1.  Determine  the  projection  upon  the  X-axis  of  a  segment,  5  feet  long,  of 
a  line  which  makes  with  the  X-axis  an  angle  of  30°. 

2.  Determine  the  projection  of  the  same  line  upon  the  F-axis. 

3.  Determine  the  projections  upon  the  X-  and  F-axes  of  a  segment,  7  feet 
long,  of  a  line  parallel  to  the  F-axis. 

4.  Show  that  the  sum  of  the  projections  upon  the  X-axis  of  the  sides  of 
the  square  of  Figure  20  is  equal  to  zero.  Show  that  the  same  result  holds 
for  the  projections  upon  the  F-axis. 

5.  (See  Fig.  21.)  Show  that  the  projection  of  AC  upon  the  X-axis  (F-axis) 
is  equal  to  the  sum  of  the  projections  of  AB  and  BC  upon  the  X-axis  (F-axis). 


-s-X 


Fig.  20 


Fig.  21 


48.  Applications.  Numerous  problems  in  different  fields  of 
science  can  be  solved  by  means  of  the  methods  developed  in  the 
preceding  paragraphs.  To  solve  such  problems  there  is  required, 
however,  besides  a  knowledge  of  trigonometry,  an  understanding 
of  the  technical  terms  used  in  those  different  fields.  We  explain 
below  a  few  technical  terms  which  will  be  used  in  the  exercises  of 
the  following  paragraph  and  in  Chapter  VII.  For  further  appli- 
cations the  student  is  referred  to  books  on  surveying,  navigation, 
astronomy,  artillery,  etc. 

The  angle  made  by  the  line  along  which  an  object  is  sighted 
with  a  horizontal  line  through  the  point  of  observation  and  in 
the  same  vertical  plane  as  the  line  of  sight,  is  called  the  angle  of 
elevation  or  the  angle  of  depression  of  the  object,  according  as  the 
object  is  higher  or  lower  than  the  point  of  observation. 

The  surveyor  frequently  designates  the  direction  of  a  line  by 
means  of  its  bearing,  i.e.,  the  acute  angle  which  the  line  makes 
with  a  N.  S.  line  through  the  point  of  observation,  indicating  at 
the  same  time  whether  the  line  runs  to  the  east  or  the  west  of  the 


EXERCISES 


39 


N.  S.  line.     The  bearings  of  the  lines  OA,  OB  and  OC  in  Figure 
22  are  denoted  as  N  23°  E   (read  23  degrees  East  of  North), 


Fig.  22 


Fig.  23 


N  47°  W  (read  47  degrees  West  of  North)  and  S  18°  W  (read 
18  degrees  West  of  South)  respectively. 

The  points  of  the  compass,  as  used  by  the  navigator,  are  indi- 
cated on  Figure  23. 

49.   Exercises. 

1.  The  angle  of  elevation  of  the  top  of  a  mountain  from  a  point  A,  situ- 
ated in  a  plane  1500  feet  below  the  top,  is  19°  27'.  Determine  the  distance 
from  A  to  the  top  in  an  air  line;  also  the  horizontal  distance  from  A  to  the 
foot  of  the  mountain. 

2.  From  a  lightship  L,  at  a  distance  of  500  feet  from  a  point  ^4  on  shore, 
the  angle  of  elevation  of  a  water  tower  vertically  above  A  is  28°  33'.  Deter- 
mine the  height  of  the  tower  above  the  level  of  the  ship. 

3.  The  angle  of  depression  of  a  point  P  as  seen  from  an  airship  1S00  feet 
above  the  ground  is  (33°  27'.  What  is  the  straight  line  distance  of  the  air- 
ship from  P? 

4.  An  observation  tower  T  is  40  feet  high  and  stands  20  feet  from  the 
edge  of  a  vertical  cliff,  whose  top  is  400  feet  above  sea  level.  A  ship  S  is  in 
the  vertical  plane  through  T  and  on  a  line  at  right  angles  to  the  shore  line; 
its  angle  of  depression  from  T  is  9°  2)5'  10".      How  far  is  <S  from  the  shore? 

5.  From  a  point  .4  on  the  edge  of  a  stream  and  5  feet  above  the  ground, 
the  angle  of  elevation  of  the  top  of  a  tree  straight  across  the  stream  is  20°  13'. 
The  height  of  the  tree  above  the  ground  is  estimated  to  be  15  feet.  How 
wide  is  the  stream  at  that  point? 


40  EXERCISES 

6.  A  vessel  is  observed  directly  south  from  a  lighthouse  L  and  S  43°  27' 
W  from  a  lighthouse  M  known  to  be  50  miles  due  East  from  L.  What  are 
the  distances  of  the  vessel  from  each  of  the  lighthouses? 

7.  From  points  P  and  Q,  150  feet  apart  and  both  lying  in  the  same  ver- 
tical plane  through  a  spire  S,  the  angles  of  elevation  of  the  spire  are  observed 
to  be  12°  13'  and  28°  36'  respectively.  How  high  is  the  spire  if  P  and  Q  are 
on  the  same  side  of  the  spire;   if  P  and  Q  are  on  opposite  sides  of  the  spire? 

8.  Two  points,  P  and  Q,  known  to  be  a  mile  apart  on  a  level  road  which 
lies  in  a  vertical  plane  through  the  top  of  a  mountain  A  are  observed  from  A. 
The  angles  of  depression  of  P  and  Q  are  23°  17'  and  32°  27'  respectively. 
How  high  is  the  mountain  top  above  the  road? 

9.  The  bearing  of  the  line  CA  is  N  12°  E;  the  bearing  of  the  line  CB  is 
S  78°  E,  while  A  bears  N  15°  W  from  B.  The  distance  CB  is  known  to  be 
257  feet.     Determine  the  distances  from  A  to  B  and  to  C. 

10.  From  a  point  B  on  one  side  of  a  stream  and  5  feet  above  the  ground, 
the  angle  of  elevation  of  a  point  P  directly  across  on  the  opposite  shore  is 
found  to  be  34°  13'.  A  level  line  BC,  100  feet  long  is  laid  off  at  right  angles 
to  the  stream,  and  from  C  the  angle  of  elevation  of  P  is  found  to  be  20°  43'. 
Determine  the  height  of  P  and  the  width  of  the  stream. 

11.  A  railroad  track  consists  of  a  horizontal  piece  followed  by  a  down- 
ward grade  4  miles  long,  making  an  angle  of  12°  with  the  horizontal.  It  is 
proposed  to  replace  the  12°  grade  by  a  4°  grade.  How  much  of  the  hori- 
zontal track  must  be  removed  to  accomplish  this  change? 

12.  How  much  track  mileage  would  be  saved  by  the  change  proposed  in 
Problem  11? 

13.  The  Washington  monument  is  555  feet  high.  From  a  point  P  on  a 
level  with  the  base,  the  angle  of  elevation  of  the  top  is  60°.  How  far  is  P 
from  the  bottom  of  the  monument?     How  far  is  P  from  the  top? 

14.  Raising  the  ridgepole  of  a  roof  3  feet,  changes  the  angle  under  which 
the  rafters  slope  from  32c  to  40°.  How  high  was  the  roof  originally,  and  how 
long  are  the  tie  beams? 

15.  A  lightship  is  observed  due  West  at  10  a.m.  and  X40°W  at  2  p.m.  by 
a  vessel  that  is  traveling  due  South  at  20  miles  per  hour.  How  far  is  the  vessel 
from  the  lightship  at  10  a.m.,  at  noon,  at  2  p.m.? 


CHAPTER  V 
THE    GRAPHS    OF    THE    TRIGONOMETRIC    FUNCTIONS 

50.  Graphs  of  sin  8  and  cos  8.  Many  properties  of  the  trigo- 
nometric functions  appear  in  a  new  light  if  we  consider  their 
graphical  representation.  In  this  chapter  we  shall  be  concerned 
principally  with  the  graphical  representations  (usually  called  the 
graphs)  of  the  functions  sin  8  and  cos  8  and  of  the  sine  and  cosine 
functions  of  expressions  of  the  form  ad  +  b,  i.e.,  of  sin  (ad  +  b) 
and  cos  (ad  +  b):  For  their  construction  we  make  use  of  the 
graphical  representation  of  pairs  of  numbers  used  in  Chapter  I. 

Angles  8  measured  by  an  arbitrary  unit  will  be  represented  by 
distances  along  the  X-axis  (abscissae) ;  with  each  abscissa  will 
be  associated  as  ordinate  a  line  representing  the  value  of  that 
function  of  the  angle  8  whose  graph  we  wish  to  obtain.  The 
determination  of  these  ordinates  can  be  carried  out  graphically 
by  the  following  simple  device. 

After  having  established  the  unit  of  measurement  on  the  X- 
and   F-axis,  we  describe  a  circle  whose  radius  is  equal  to  this 


A      I 


O  U    1 

-e- H 


Fig.  24 


->>-X 


unit  and  whose  center  is  on  the  X-axis  (see  Fig.  24).  At  the 
center  of  this  circle  we  construct  an  angle  8  whose  initial  side  lies 
in  the  positive  direction  along  the  X-axis.  From  the  point  P, 
where  the  terminal  side  of  8  meets  the  circle,  we  drop  a  perpen- 

41 


42 


EXAMPLES  OF  GRAPHS 


dicular  PQ  upon  the  X-axis.  Since  for  P  the  radius  vector  is 
equal  to  unity,  the  sine  of  angle  d  is  measured  in  magnitude  and 
direction  by  the  ordinate  of  P,  i.e.,  by  the  line  QP.  Through  P  we 
draw  a  line  PP',  parallel  to  the  X-axis.  We  then  lay  off  on  the 
X-axis  a  distance  OA,  representing  the  angle  6  in  magnitude  and 
in  sense,  through  A  we  draw  a  line  A  A'  parallel  to  the  F-axis, 
meeting  the  line  PP'  in  B;  this  point  B  is  then  a  point  on  the 
graph  of  sin  0. 

If  the  angle  0  be  laid  off  with  its  initial  side  in  the  positive 
direction  along  the  F-axis  and  on  the  vertical  diameter  of  the 


Fig.  25 

circle  (see  Fig.  25),  we  obtain  an  ordinate  QP  which  measures  in 
magnitude  and  direction  the  cosine  of  0. 

51.   Examples  of  graphs. 

(a)  Construct  the  graph  of  sin  0  for  6  varying  from  0°  to  360°  (see 
Fig.  26). 


>Xa 


, 

r 

>"i 

>2 

■"-> 

\ 

v~ 

/-V-W 

///\P2 

N 

rww 

////>, 

~7Bi 

1\ 

^^ — Si 

/fl 

\ 

[^-^^M 

ok, 

A{ 

WW/ 

\\\y 

— 

? 

Fig.  26 


At  the  center  of  the  auxiliary  circle  we  lay  off  angles  of  0°,  15°, 
30°,  .  .  .  330°.  345°,  360°,  with  initial  sides  in  the  positive  direc- 
tion along  the  X-axis  and  with  terminal  sides  cutting  the  circle 


EXAMPLES  OF  GRAPHS 


43 


in  points  P0,  Pi,  P2,  etc.  On  the  X-axis  we  lay  off  distances  0Aa 
=  0,  OAi  =  ir/ 12,  OA2  =  x/6,  etc.,  representing  the  radian  meas- 
urements of  these  angles.  Through  P0,  Pi,  P2,  etc.  we  draw  lines 
PoX0,  P1X1,  P2X2,  etc.  parallel  to  the  X-axis;  through  A0,  Ah  A2, 
etc.  we  draw  lines  A0Y0,  A\Y\,  A2Y2,  etc.  parallel  to  the  F-axis. 
The  intersections  B0,  B1}  B2,  etc.  of  the  lines  A0Y0  and  PoX0, 
A\Yi  and  P1X1,  A2Y2  and  P2X2,  etc.  respectively,  are  points  whose 
:r-coordinates  measure  angles  and  whose  ^-coordinates  measure 
the  sines  of  these  angles.  A  smooth  curve  drawn  through  the 
points  Bo,  Bi,  B2,  etc.  is  the  graph  of  the  function  sin  6  for  6  vary- 
ing from  0°  to  360°. 

(6)  Construct  the  graph  of  cos  6  for  6  varying  from  0°  to  360°  (see 
Fig.  27). 


r 
* 

\ 

*2 

X0l 

/  /-s> 

ui?>. 

a2S\- 

i//sS\ 

p^§ 

jj£^~~\ 

i^^? 

%^r~~^  I 

oki-i1 

y-v/y 

\\V\/ 

\\y 

i  ^ 

, 

~i 

— 

h- 

■ 

Fig.  27 


We  construct  at  the  center  of  the  auxiliary  circle  angles  of  0°, 
15°,  30°,  .  .  .  345°,  360°  with  initial  sides  in  the  positive  direc- 
tion along  the  Y-axis  and  with  terminal  sides  cutting  the  circle 
in  points  Q0,  Qu  Q2,  etc.  On  the  X-axis  we  determine  the  points 
A0  Ah  A2,  etc.  as  in  example  (a).  Lines  Q0X0,  Q1X1,  Q2X2,  etc. 
parallel  to  the  X-axis  will  meet  lines  A0Y0,  AxYh  A2Y2,  etc. 
parallel  to  the  F-axis  in  points  B0i  Bu  B2,  etc.  whose  ^-coordi- 
nates measure  angles  and  whose  ^-coordinates  measure  the  cosines 
of  these  angles.  A  smooth  curve  drawn  through  these  points 
Bo,  7>i,  B2,  etc.  is  the  graph  of  the  function  cos  6  for  6  varying 
from  0°  to  360°. 

(c)  Construct  the  graph  of  sin  (2  9  —  30°)  for  6  varying  from 
0°  to  360°.  We  construct  as  in  example  (a)  the  lines  P0X0,  P1X1, 
P2X2,  etc.  and  A0Y0  A^Yh  A2Y2,  etc.  By  means  of  these  lines 
we  determine  a  point  B0,  whose  ^'-coordinate  measures  an  angle 
of  0°  and  whose  ^-coordinate  measures  the  sine  of  2  •  0°  —  30°, 


44 


EXERCISES 


i.e.,  the  sine  of  —30°;  a  point  Bh  whose  ^-coordinate  measures 
an  angle  of  15°,  and  whose  y-coordinate  measures  the  sine  of 
2  •  15°  —  30°,  i.e.,  the  sine  of  0°;  a  point  B2,  whose  z-coordinate 
measures  an  angle  of  30°  and  whose  ^/-coordinate  measures  the 
sine  of  2  •  30°  —  30°,  i.e.,  the  sine  of  30°,  etc.  A  smooth  curve 
through  the  points  B0,  B1}  B2,  etc.  is  the  required  graph. 

(d)  Construct  the  graph  of  cos  (0/3  +  45°)  for  6  varying  from 
0°  to  360°  (see  Fig.  28). 


Q,QJ 

3 
i 

Qo 

r 
] 

* 

y> 

— 5To  v. 

«2^" 

TF1 

/\\ 

/  /\ 

'x?, 

/\\\ 

/ 

WW 

l^^a 

Ut£Z^~~\ 

■+-JC 

1^~— --J 

AXA2 

\///l  \\ 

1 

\// 1  \  \\  y 

i 

H 

Fig.  28 

We  construct,  as  in  example  (b)  the  lines  QoXq,  QiX\,  Q2X2, 
etc.  and  A0Y0,  AiYh  A2Y2,  etc.  By  means  of  these  lines  we 
determine  a  point  B0,  whose  ^-coordinate  measures  an  angle  of 
0°  and  whose  ^/-coordinate  measures  the  cosine  of  0°/3  +  45°, 
i.e.,  cos  45°;  a  point  Bx  whose  aj-coordinate  measures  an  angle 
of  15°  and  whose  ^-coordinate  measures  the  cosine  of  15°/3 
+  45°,  i.e.,  cos  50°;  a  point  B2,  whose  x-coordinate  measures  an 
angle  of  30°,  and  whose1  y-coordinate  measures  the  cosine  of 
30°/3  +  45°,  i.e.,  cos  55°,  etc.  A  smooth  curve1  drawn  through 
the  points  Bo,  Bh  B2,  etc.  is  the  required  graph. 


52.    Exercises. 

Construct  the  graphs  of  the  following  functions,  for  9  varying  from  0°  to 


360°: 

1.  cos  2  9. 

2.  cos  (9/2. 

3.  sin  2  9. 

4.  sin  9/2. 

5.  2  cos«. 

6.  2  sin  9. 

7.  \  sin  9. 

8.  |  cos  9. 


9.    sin  3  9. 
10.    3  sin  9. 
3  cos  9  3. 
cos  3  9. 
\  cos  3  9. 
3  sin  9  3. 
I  sin  3  9. 
I  cos  3  a. 


11. 
12. 
13. 
14. 
15. 
16. 


17. 

cos  (— e). 

18. 

sin  (0  +  90°). 

19. 

cos  1.9  -  ISO3) 

20. 

cos  (9  +  180°) 

21. 

cos  (6  -  90°). 

22. 

sin  (0  -  00°). 

23. 

cos  (0  +  270°) 

24. 

sin  '9  -  270°). 

OPERATIONS  ON   GRAPHS  45 

53.  Operations  on  graphs.  The  graph  of  the  function  sin  3  0 
associates  with  any  particular  value  of  0  the  same  ordinate  that 
the  graph  of  sin  0  associates  with  3  0,  a  point  three  times  as  far 
from  the  origin  as  0.  Hence  to  every  point  A  on  the  graph  of 
the  function  sin  0  there  corresponds  a  point  B  on  the  graph  of  the 
function  sin  3  9,  three  times  as  near  to  the  F-axis  as  the  point  A. 
It  follows  from  this  that  the  graph  of  the  function  sin  3  9  ma\' 
be  obtained  by  contracting  the  graph  of  the  function  sin  0  in 
the  direction  of  the  X-axis  in  the  ratio  3:1.  If  we  apply  these 
same  considerations  to  the  general  case,  we  obtain  the  following 
theorem : 

Theorem  I.  The  graphs  of  the  functions  sin  ad,  cos  ad  may  be 
obtained  by  contracting  the  graphs  of  the  functions  sin  9  and  cos  9 
respectively  towards  the  Y-axis  in  the  ratio  a  :  1,  if  a  is  a  positive 
rational  number. 

Here  it  is  to  be  understood  that  if  a  <  1,  the  contraction 
becomes  an  enlargement  in  the  ratio  1  :  1/a. 

The  graph  of  sin  (  —  9)  associates  with  each  value  of  0  the 
same  ordinate  that  the  graph  of  sin  0  associates  with  —  0.  Hence 
to  every  point  A  on  the  graph  of  sin  0  there  corresponds  a  point  B 
on  the  graph  of  sin  (—8)  obtained  by  reflecting  A  in  the  F-axis 
as  a  mirror.*     This  leads  to  the  following  theorem: 

Theorem  II.  The  graphs  of  the  functions  sin  (—  9),  cos  (— 0)  may 
be  obtained  by  reflecting  the  g  aphs  of  the  functions  sin  9  and  cos  9 
respectively  in  the  Y-axis  as  a  mirror. 

The  graph  of  the  function  sin  (2  9  +  180°)  associates  with  any 
particular  value  of  9  the  same  ordinate  that  the  graph  of  sin  2  9 
associates  with  9  -f-  90°,  a  point  on  the  X-axis  90°  farther  to  the 
right  than  9.  Hence,  to  every  point  A  on  the  graph  of  sin  2  9  there 
will  correspond  a  point  B  on  the  graph  of  sin  (2  9  -\-  180°),  a  dis- 
tance of  90°  to  the  left  of  A .  It  follows  from  this  that  the  graph 
of  sin  (2  9  +  180°)  can  be  obtained  by  translating  the  graph  of 
sin  2  9  in  the  direction  of  the  negative  X-axis  a  distance  of  90°. 
In  the  same  way,  we  see  that  the  graph  of  sin  (2  9  —  180°)  may 
be  obtained  by  translating  the  graph  of  sin  2  9  in  the  direction  of 
the  positive  X-axis  a  distance  of  90°. 

*  11  is  said  to  be  the  reflection  of  .1  in  the  F-axis  as  a  mirror  if  the  V-axis 
is  the  perpendicular  bisector  of  the  line  AB. 


46 


OPERATIONS  ON   GRAPHS 


If  we  carry  through  the  same  considerations  for  the  general 
case,  we  obtain  the  following  theorem: 

Theorem  III.  The  graphs  of  the  functions  sin  (ad  +  b),  cos  (ad  +  b) 
may  be  obtained  by  translating  the  graphs  of  sin  ad  and  cos  ad  respec- 
tively a  distance  of  |  b/a  \,*  in  the  direction  of  the  negative  X-axis,  if  b/a 
is  positive;  in  the  direction  of  the  positive  X-axis,  if  b/a  is  negative. 

Theorems  I,  II  and  III  may  be  used  to  obtain  the  graphs  of  the 
functions  sin  {ad  +  b)  and  cos  (ad  +  b),  where  a  and  b  are  arbi- 
trary rational  numbers  by  translating,  reflecting  and  contracting 
(enlarging)  the  graphs  of  the  functions  sin  6  and  cos  0  respectively. 
This  will  now  be  illustrated  by  examples. 

(1)  To  obtain  the  graph  of  the  function  sin  (3  0  +  30°),  we  first 
contract  the  graph  of  the  function  sin  0  horizontally  in  the  ratio 
3:1,  obtaining  in  this  way  the  graph  of  the  function  sin  3  0 
(see  Fig.  29),  in  virtue  of  Theorem  I. 


*x 


Fig.  29 


We  now  translate  the  graph  of  sin  3  0  a  distance  of  10°  to  the 
left  and  obtain  in  this  way,  in  virtue  of  Theorem  III,  the  graph 

of  sin  (3  [6  +  10°])  =  sin  (3  0  +  30°). 


+~X 


Fig.  30 

(2)  To  construct  the  graph  of  the  function  sin  (90°  —  6),  we  re- 
flect the  graph  of  sin  0  in  the  F-axis  as  a  mirror,  which  gives  us 
the  graph  of  sin  {  —  d),  in  virtue  of  Theorem  II  (see  Fig.  30). 

*  The  symbol  |  b/a  |  designates  the  numerical  value  of  b/a.  Thus,  if  a—  —  1, 
b  =  90°,  |  b/a  |  =  |  -90°  |  =  90°;  similarly,  |  -4  |  =  4,  |  -f  |  =|. 


APPLICATIONS  OF  GRAPHS  47 

Translating  this  latter  graph  a  distance  of  90°  to  the  right,  we 
shall  obtain,  in  virtue  of  Theorem  III,  the  graph  of  sin  (  —  [6  —  90°]) 
=  sin  (90°  -  6). 

54.  Exercises. 

Construct  the  graphs  of  the  following  functions: 

1.  cos  (90°  +  0).  7.  cos  (0  +  180°).  13.  cos  (3  0  -  60°). 

2.  sin  (90°  -  0).  8.  cos  (180°  -  0).  14.  sin  (0  -  180°). 

3.  cos  (90°  -  6).  9.  sin  (270°  +  0).  15.  sin  (3  0  +  45°). 

4.  sin  (0  -  90°).  10.  sin  (270°  -  0).  16.  cos  (90°  -  2  d). 

5.  sin  (90° +  0).  11.  sin  (2  0 +60°).  17.  sin  (60° +  3  0). 

6.  cos  (0  -  90°).  12.  cos  (2  0  -  90°).  18.  cos  (180°  -  4  0). 

55.  Applications  of  graphs.  The  graphs  of  the  trigonometric 
functions  sin  (ad  +  b)  and  cos  (ad  +  b),  which  we  have  learned 
to  construct  in  the  foregoing  paragraphs,  will  now  be  used  to 
illustrate  and  verify  some  of  the  important  properties  of  the  func- 
tions sin  d  and  cos  d. 

(1)  The  functions  sin  ax  and  cos  ax  are  periodic  functions 
whose  period  is  equal  to  2  t/  |  a  (  ,  where  a  represents  any  rational 
number.     (See  footnote  on  p.  46  and  26.) 

(2)  The  graphs  of  the  functions  sin  (—d)  and  sin  6  are  symmet- 
ric with  respect  to  the  X-axis,  i.e.,  for  any  value  of  d,  the  corre- 
sponding values  of  these  functions  are  equal  but  opposite  in  sign. 
We  conclude  therefore: 

(1)  sin  (-0)  =  -sine,    for  every  value  of  6. 

The  graphs  of  the  functions  cos  (—  6)  and  cos  0  coincide,  so 
that  we  have: 

(2)  cos  (-0)  =  cos0,    for  every  value  of  0. 

Formula  (1)  shows  an  analogy  between  the  sine  function  and  an 
odd  power  of  a  variable;  for  we  know  that,  e.g.,  (  —  d)lh  =  —  015. 
On  the  other  hand  formula  (2)  shows  an  analogy  between  the 
cosine  function  and  an  even  power,  for,  e.g.,  (  —  0)lfi  =  6ir'.  On 
account  of  this  analogy,  formulae  (1)  and  (2)  are  frequently 
expressed  in  the  form  "  the  sine  is  an  odd  function,"  and  "  the 
cosine  is  an  even  function." 

(3)  The  graph  of  the  function  sin  (90°  —  6)  coincides  with  the 
graph  of  cos  6;  from  this  we  conclude: 

(3)  sin  (90°  -  6)  -  cos  6,  for  every  value  of  6.     (See  19,  Ex.  1.) 


48 


GRAPHS  OF  TAN  0  AND  COT  8 


Formula  (3)  as  well  as  most  of  those  which  appear  in  the  exer- 
cises below  can  also  be  obtained  by  the  use  of  Theorem  I  in  Chap- 
ter II.  The  advantage  gained  by  deriving  them  by  the  present 
method  lies  in  the  fact  that  it  emphasizes  more  sharply  the  fact 
that  these  relations  hold  for  every  value  of  the  variable,  i.e.,  that 
they  are  properties  of  the  trigonometric  functions. 


56.   Exercises. 

Prove  graphically: 

1.  sin  (180°  -0)  =  sin(9. 

2.  sin  (0  -  180°)  =  -sin0. 

3.  sin  ((9  +  180°)  =  -sin  0. 

4.  cos  (90°  +  6)  =  -sin0. 

5.  cos  (0  -  90°)  =  sin  0. 

6.  cos  (180°  +  6)  =  -cos  0. 

7.  sin  (90°  +6)  =  cos  0. 

8.  sin(0  -  90°)  =  -cos0. 

9.  cos  (90°  -  0)  =  sin  0. 
10.   cos  (180°  -  0)  =  -cos0. 


11.  cos  (0  -  180°)  =  -cos0. 

12.  sin  (0  +  270°)  =  -cos0. 

13.  cos  (0  +  270°)  =  sin  0. 

14.  sin  (270°  -  0)  =  -cos0. 

15.  cos  (270°  -  0)  =  -sin  0. 

16.  cos  (0  -  270°)  =  -sin  0. 

17.  sin  (0  -  270°)  =  cos  0. 

18.  sin  (0  -  45°)  +  sin  (0  +  45°)  =  V2  •  sin  9. 

19.  sin  (0  +  45°)  -  sin  (0  -  45°)  =  V2  •  cos  0. 

20.  sin  0  +  cos  0  =  V2  •  sin  (0  +  45°). 


57.  Graphs  of  tan  6  and  cot  6.  We  proceed  now  to  a  brief 
consideration  of  the  graphs  of  he  functions  tan  9  and  cot  6.  For 
this  purpose  we  begin  by  developing  a  graphical  method  for 
determining  an  ordinate  which  measures  the  tangent  of  a  given 
angle  6,  analogous  to  the  method  developed  for  the  sine  in  50. 


*~x 


Fig.  31 

We  return  to  the  auxiliary  circle  used  in  50  and  51  and  draw  a 
line  TT'  tangent  to  this  circle  at  the  right  hand  extremity  of 
its  horizontal  diameter  (see  Fig.  31).     At  the  center  of  the  circle 


GRAPHS  OF  TAN  d  AND  COT  6 


49 


we  draw  an  angle  9  whose  initial  side  lies  in  the  positive  direction 
along  the  X-axis.  From  the  point  S,  where  the  terminal  side 
of  9  or,  for  angles  in  II  and  III,  the  terminal  side  of  9  produced 
through  the  origin,  cuts  the  tangent  line  TT',  we  draw  a  perpen- 
dicular SR  upon  the  X-axis.  Since  for  S  the  abscissa  is  equal 
to  unity  the  tangent  of  9  is  measured  by  the  ordinate  of  S,  i.e.,  by 
the  line  RS.  Through  S  we  draw  a  line  SS'  parallel  to  the  X-axis. 
Furthermore,  we  lay  off  on  the  X-axis,  as  in  50,  a  distance  OA, 
representing  the  angle  9;  through  A  we  draw  a  line  A  A'  parallel 
to  the  F-axis,  meeting  the  line  SS'  in  B;  this  point  B  is  then  a 
point  on  the  graph  of  tan  9. 

Now,  following  the  general  method  explained  in  51,  we  readily 
construct  the  graph  of  the  function  tan  (ad  +  6). 

To  obtain  an  ordinate  measuring  the  cotangent  of  9,  we  draw  a 
line  UU'  tangent  to  the  circle  at  the  left  hand  extremity  of  its 
horizontal  diameter.  We  lay  off  the  angle  9  at  the  center  of  the 
circle,  with  its  initial  side  in  the  positive  direction  along  the 
F-axis,  and  determine  the  point  S  where  the  terminal  side  of  9,  or 
the  terminal  side  produced  through  the  origin,  meets  this  tangent 
line;  from  S  we  drop  a  perpendicular  SR  upon  the  X-axis.  It  is 
clear  (see  Fig.  32)  that  if  we  consider  this  position  of  9  as  the 


Fig.  32 


standard  position,  the  ordinate  for  the  point  S  is  equal  to  unity, 
so  that  the  cotangent  of  6  is  measured  by  the  abscissa,  i.e.,  by 
the  line  RS,  no  matter  how  large  the  angle  9  may  be.  Through 
S  we  draw  a  line  SS'  parallel  to  the  X-axis;  we  lay  off,  as  before, 
a  distance  OA  on  the  X-axis,  measuring  the  angle  9,  draw  the 
line  A  A'  parallel  to  the   F-axis  and  determine  the  point   B,  in 


50  GRAPHS  OF  TAN  AND  COT   (ad+b) 

which  the  lines  SS'  and  A  A'  meet.  This  point  B  is  a  point  on 
the  graph  of  cot  8.  The  graph  of  the  general  function  cot  (ad  -f-  6) 
is  now  constructed  by  the  methods  used  in  51. 

58.  Exercises. 

Construct  the  graphs  of  the  following  functions,  for  0  varying  from  0°  to 
360°: 

1.  tan  2  9.  7.  tan  0/3.  13.  2  tan  9. 

2.  cote/2.  8.  cot  3  9.  14.  3  tan  9/3. 

3.  tan  (9  +  90°).  9.  cot  (9  +  270°).  15.  tan  (2  6  +  90°). 

4.  cot  (6  -  90°).  10.  tan  (0  -  270°).  16.  cot  {2  9  -  180°). 
6.  tan  (9  +  180°).  11.  tan  (180°  -  6).  17.  cot  (180°  +  2  9). 
6.  cot  (9  +  180°).  12.  cot  (180°  -  9).  18.  tan  (180°  -  2  9). 

59.  Graphs  of  tan  (ad  +  b)  and  cot  (ad  +  b).  Mere  repetition 
of  the  arguments  of  53  will  enable  us  at  once  to  establish  the 
following  theorems,  analogous  to  Theorems  I,  II,  III  of  53. 

Theorem  la.  The  graphs  of  the  functions  tan  ad,  cot  a9  may  be 
obtained  by  contracting  the  graphs  of  the  functions  tan  9  and  cot  9 
respectively  towards  the  Y-axis  in  the  ratio  a  :  1,  if  a  is  a  positive 
rational  number;  if  a  <  1,  the  contraction  becomes  an  enlargement 
in  the  ratio  1 : 1/a. 

Theorem  Ha.  The  graphs  of  the  functions  tan  (-0),  cot  (—9)  may 
be  obtained  by  reflecting  the  graphs  of  the  functions  tan  e  and  cot  e 
respectively  in  the  Y-axis  as  a  mirror. 

Theorem  Ilia.  The  graphs  of  the  functions  tan  (a9  +  b),  cot 
(ad  +  b)  may  be  obtained  by  translating  the  graphs  of  the  functions 
tan  a9  and  cot  a9  respectively  a  distance  of  |  b/a  \  ,  in  the  direction  of 
the  negative  X-axis,  if  b/a  is  positive;  in  the  direction  of  the  positive 
X-axis,  if  b/a  is  negative. 

These  theorems  enable  us  to  obtain  the  graphs  of  functions  of 
the  form  tan  (ad  +  /;)  and  cot  (ad  +  b),  where  a  and  b  are  arbitrary- 
rational  numbers,  by  translating,  reflecting  and  contracting  (en- 
larging) the  graphs  of  tan  6  and  cot  6  respectively. 

(a)  To  construct  the  graph  of  tan  (2  d  +  45°),  we  first  contract 
the  graph  of  tan  d  horizontally  in  the  ratio  2  :  1,  so  as  to  obtain 


APPLICATIONS 


51 


the  graph  of  tan  2  d.     Then  we  translate  the  latter  graph  a  dis- 
tance of  22|°  in  the  direction  of  the  negative  X-axis.     (See  Fig.  33.) 


Fig.  33 


(b)  To  construct  the  graph  of  cot  (180°  —26),  we  reflect  the 
graph  of  cot  d  in  the  F-axis  as  a  mirror,  and  contract  the  latter  in 
the  ratio  2  :  1  in  the  direction  of  the  X-axis,  obtaining  in  this 
way  the  graph  of  cot  (—2  6).  If  we  now  translate  this  graph  a 
distance  of  90°  in  the  direction  of  the  positive  X-axis,  we  obtain 
the  graph  of  the  function  cot  (-2  [6  -  90°])  =  cot  (180°  -  2  6), 
which  proves  to  be  identical  with  the  graph  of  cot  (  —  2  6).  (See 
Fig.  34.) 


Fig.  34 

60.  Applications.  The  graphs  of  the  functions  tan  (ad  +  b) 
and  cot  (ad  -f-  b)  may  now  be  used  to  derive  some  of  the  prop- 
erties of  the  functions  tan  6  and  cot  6,  by  the  method  explained 
in  55.     We  obtain  in  this  way  the  following  results: 

(1)  the  functions  tan  ad  and  cot  ad  are  periodic  functions  whose 
period  is  w/\  a  |. 


52 


GRAPHS  OF  SEC  0  AND   COSEC  0 


It  is  to  be  observed  that  the  period  of  the  tangent  and  cotangent 
functions  is  half  as  large  as  that  of  the  corresponding  sine  and 
cosine  functions. 

(2)  the  tangent  of  ir/2  and  3  w/2,  and  the  cotangent  of  0  and  w 
do  not  exist  (see  Theorem  IV,  Chapter  II).     Moreover 


Lim  tan  0  =  +  °c 

0— >90°- 

Lim  cot  0  =  —  oo . 

0^>18O°- 


Lim  tan  0  =  —  oo . 

0  — >  90°+ 

Lim  cot  0  =  +  oo .     (See  24, 25.) 

fl— >.180°+ 


(3)  the  tangent  and  cotangent  are  odd  functions,  i.e., 

tan  (— 0)  =  —  tan  0,        cot  (— 0)  =  —cote. 

(4)  tan  (180°  -  0)  =  -  tan  0,  etc. 

tan  (90°  —  0)  =  cot  0,  etc.     (See  exercises  below.) 


61.   Exercises. 

Construct  the  graphs  of  the  following  functions: 

1.  tan  (90°  +  0).  5.    tan  2  0. 

2.  eot(-0).  6.   tan  (180° +  0). 

3.  tan  (90°  -  0).  7.   cot  (0  -  180°). 

4.  cot  (90°  + 0).  8.    cot  0/2. 


Prove  graphically: 

13.  tan  (9  +  180°)  =  tan  0. 

14.  cot  (0  +  180°)  =  cot  0. 

15.  tan  (90°  +  0)  =  -cot0. 

16.  cot  (90°  -  0)  =  tan  0. 

17.  tan  (0  +  270°)  =  -cot  6 

18.  cot  (90°  +0)  =  -tan0. 


9.  tan  (3  0  -  90°). 

10.  tan  (0/2  +  90°). 

11.  cot  (2  0  -  90°). 

12.  tan  (90°  -  2  0). 


19.  tan  (0  -  180°)  =  tan  0. 

20.  cot  (270° +  0)  =  -tan  6 

21.  tan  (270°  -  0)  =  cot  0. 

22.  cot  (0  -  270°)  =  -tan  6 

23.  tan  (45°  +  0)  =  cot  f4oc 

24.  tan  (45°  -  0)  =  cot  (45° 


62.  Graphs  of  sec0  and  cosec  0.  For  the  sake  of  complete- 
ness, we  add  a  method  for  the  construction  of  the  graphs  of  the 
functions  sec  0  and  cosec  0.  For  the  former,  we  use  the  diagram 
of  Figure  31.  Since  for  the  point  S,  the  abscissa  is  equal  to  unity, 
the  secant  of  0  is  measured  by  the  radius  vector  MS.  If  we 
describe  an  arc  with  M  as  center  and  MS  as  radius  we  may  deter- 
mine on  the  vertical  diameter  of   the  circle  a  point  Si,  whose 


GRAPHS  OF  SEC0  AND   COSEC  6 


53 


perpendicular  distance  from  the  X-axis  is  equal  in  magnitude 
and  direction  to  sec  6*  Through  $1  we  draw  a  line  SiSi'  parallel 
to  the  X-axis  meeting  the  line  A  A'  in  a  point  B,  which  is  a  point 
on  the  graph  of  sec  6.  The  graph  is  completed  by  means  of  the 
method  of  51  (see  Fig.  35). 


Fig.  36 


For  the  graph  of  the  cosecant  function  we  use  a  diagram  similar 
to  the  one  used  for  the  cotangent  (see  Fig.  32).     The  ordinate  of  £ 

*  Here  it  is  to  be  understood  that  Si  is  to  be  determined  on  the  positive 
half  of  the  diameter,  if  S  lies  on  the  terminal  side  of  0;  on  the  negative  half  of 
the  diameter  if  S  lies  on  the  terminal  side  of  0  produeed  through  the  origin. 


54 


EXERCISES 


being  equal  to  unity,  the  cosecant  of  6  is  measured  by  the  radius 
vector  MS.  Describing  an  arc  with  center  at  M  and  radius 
equal  to  MS,  we  find  on  the  vertical  diameter  of  the  circle  a  point 
Si  whose  distance  from  the  X-axis  is  equal  in  magnitude  and 
direction  to  the  cosecant  of  6  (see  footnote  on  p.  53).  Drawing 
a  line  SiSi  parallel  to  the  X-axis,  we  determine  upon  the  line 
AA' ,  drawn  parallel  to  the  F-axis,  a  point  B  which  is  a  point  of 
the  graph  of  cosec  8.  Repeating  this  construction  for  various 
values  of  0,  as  explained  in  51,  we  obtain  the  graph  of  the  function 
cosec  9  (see  Fig.  36). 

For  the  functions  sec  8  and  cosec  8  we  can  now  develop  theorems 
analogous  to  theorems  I,  II,  III  of  53,  by  means  of  which  the 
graphs  of  the  general  functions  sec  (ad  +  b)  and  cosec  (ad  +  b) 
can  be  obtained  by  translating,  reflecting  and  contracting  (enlarg- 
ing) the  graphs  of  sec  8  and  cosec  8.  This  in  turn  enables  us  to 
prove  graphically  various  properties  of  these  functions. 


63.   Exercises. 

Construct  the  graphs  of  the  following  functions: 

1.  sec  2  0.  5.    cosec  0/2. 

2.  sec  (90°  +  0).  6.    cosec  (180°  +  0). 

3.  cosec  (—0).  7.   cosec  (0  —  90°). 

4.  sec  (-0).  8.   sec  (3  0  -  270°). 


9.  cosec  3  9. 

10.  sec  (0  +  270°). 

11.  cosec  (0/2  +  90°). 

12.  sec  (180°  -  0). 


Prove  graphically: 

13.  The  secant  function  is  an  even  function. 

14.  The  cosecant  function  is  an  odd  function. 


15.  sec  (90°  -  0)  =  cosec  0. 

16.  cosec  (90°  -  0)  =  sec  0. 

17.  sec  (180°  +  0)  =  -sec0. 

18.  cosec  (180°  +  0)  =  -  cosec  0. 

19.  sec  (90°  +  0)  =  -  cosec  0. 

20.  sec  (0  -  90°)  =  cosec  0. 

21.  cosec  (90°  +  0)  =  sec  0. 

22.  cosec  (0  -  90°)  =  -  sec  0. 


23.  sec  (180°  -0)  =  -  sec  0. 

24.  sec  (9  -  180°)  =  -  sec  0. 

25.  sec  (270°  +  0)  =  cosec  0. 

26.  cosec  (270°  +0)  =  -  sec  0. 

27.  cosec  (180°  -  0)  =  cosec  0. 

28.  cosec  (0  -  180°)  =  -  cosec  9. 

29.  sec  (0  —  270°)  =  -  cosec  0. 

30.  cosec  (0  -  270°)  =  sec  0. 


CHAPTER  VI 
THE  ADDITION  FORMULAE 

64.  A  special  case.  In  the  preceding  chapter  we  have  proved 
formulae  like  sin  (0  +  90°)  =  cos  6;  cos  (90°  -  6)  =  sin  0,  etc. 
We  next  inquire  how  the  trigonometric  functions  of  the  sum  and 
difference  of  any  two  angles  may  be  expressed  in  terms  of  the 
functions  of  these  angles;  i.e.,  we  ask  in  the  first  place  for  formulae 
for  sin  (a  +  0),  sin  (a  —  0),  cos  (a  +  0)  and  cos  (a  —  0)  in  terms 
of  sin  a,  cos  a,  sin  0  and  cos  0. 

We  begin  by  deriving  in  a  slightly  different  manner  some  of 
the  formulae  mentioned  above.  From  the  graphs  of  the  functions 
sin  (a  +  90°)  and  cos  a;  cos  (a  -f  90°)  and  sin  a,  we  have  already 
concluded,  that 

(1)  sin  (a  +  90°)  =  cos  a;  and  that  (2)  cos  (o  +  90°)  =  -sin  a. 

If  in  these  formulae  we  replace  a  by  a  —  90°,  we  obtain 
(3)  sin  a  =  cos  («  -  90°),  and  (4)  cos  a  =  -sin  (a-  90°), 

which  may  also  be  derived  directly  from  the  graphs. 

65.  Addition  formulae  for  the  sine  and  the  cosine.  We  pro- 
ceed now  to  the  general  case.  We  place  the  angle  a  in  standard 
position  and  we  bring  the  initial  side  of  0  into  coincidence  with  the 
terminal  side  of  a;  let  OA  then  be  the  terminal  side  of  0. 

The  angle  a  +  /3  is  then  in  standard  position  with  respect  to 
the  axes  OX  and  OY  (see  Fig.  37),  but  not  the  angle  /3.  In  order 
to  make  possible  the  discussion  of  the  trigonometric  ratios  of  0, 
we  introduce  as  auxiliary  axes  the  lines  OP  and  OQ.  The  posi- 
tive direction  on  OP  is  that  of  the  initial  side  of  0;  the  positive 
direction  on  OQ  is  so  determined  that  when  XOY  rotates  about 
0  until  OX  coincides  with  OP,  then  OY  will  coincide  with  OQ. 
From  this  it  follows  that 

Z  XOP  =   Z  YOQ  =  a. 

We  take  now  a  point  on  the  terminal  side  of  a  -f-  /3  (which  is  at 
the  same  time  terminal  side  of  /3)  and  construct  its  coordinates 

55 


56      ADDITION  FORMULAE  FOR  THE  SINE  AND  THE  COSINE 


x  and  y  with  respect  to  the  axes  OX  and  OY,  and  also  its  coordi- 
nates x'  and  y'  with  respect  to  the  auxiliary  axes  OP  and  OQ. 
Moreover,  we  select  the  line  OA  as  our  unit  of  measurement. 
In  this  way  we  find: 
(1)       sin  /3  =  OE/OA  =  OE,     and     cos  0  =  OD/OA  =  OD. 

Furthermore,  we  see  that 

sin  (a  +  0)  =  OC/OA  =  OC  =  Projy(M. 
In  order  to  express  sin  (a  +  /3)  in  terms  of  the  trigonometric 
ratio  of  a  and  /3  separately,  we  proceed  as  in  46  and  47,  making  use 

of  the  Corollary  of  6,  with  triangle 
OB  A.     Thus  we  find: 
(2)        sin(a  +  /3)  =  ProjyOA 

=  FrojyOD  +  FrojyDA 
=  Proj  Y0D  +  Proj  y0E. 
But,  OD  is  a  segment  of  the  direct- 
-*~x  ed  line  OP,  which  makes  with  the  Y 
axis  the  angle  a  —  90°;   hence,  by 
Theorem  II  of  45, 
Proj  yOD  =  OD  •  cos  (a  -  90°) 

=  cos  0  cos  (a  -  90°).     Why? 
And,    OE    is    a  segment   of  the 
directed  line  OQ,  which  makes  with 
OY  the  angle  a;    hence  Proj yOE  =  OE  •  cos  a  =  sin  /3  cos  a. 
Why? 

Substituting  the  last  two  results  in  (2),  we  find  that 

sin  (a  +  /3)  =  cos  /3  cos  (a  —  90°)  +  sin  j8  cos  a. 
Finally,  we  make  use  of  64  to  obtain  the  result 

sin  (a  +  /3)  =  sin  a  cos  /3  +  sin  /3  cos  a. 
To  obtain  a  formula  for  cos  (a  +  /3)  we  proceed  in  the  same  way, 
projecting  on  the  X-axis  instead  of  on  the  F-axis.     This  leads  to 
the  following  development: 

cos  (a  +  j8)  =  OP/0.4  =  OB  =  Proj^Oyl 

=  Proj^OP  +  ProjzrM   -  Proj  ^07)  +  Proj*0P 
=  OD  cos  a  +  OP  cos  (a  +  90°) 
=  cos  /S  cos  a  +  sin  /3  cos  (a  +  90°)  =  cos  a  cos  /3  — 
sin  a  sin  jS 
Thus  we  have  obtained  the  following  important  result: 
cos  (a  +  /3)  =  cos  a  cos  /3  —  sin  a  sin  13, 


(3) 


sin  (a  +  $)  =  sin  a  cos  (3  +  cos  a  sin  /3; 


EXERCISES  57 

i.e.,  the  cosine  of  the  sum  of  two  angles  is  equal  to  the  product  of  the 
cosines  of  these  angles  diminished  by  the  product  of  their  sines; 
the  sine  of  the  sum  of  two  angles  is  equal  to  the  product  of  the  sine 
of  one  by  the  cosine  of  the  other  plus  the  product  of  the  sine  of  the 
other  by  the  cosine  of  the  first. 

Formulae  (3)  are  known  as  the  addition  formulae  for  the  sine 
and  cosine  functions.  The  proof  given  here  is  entirely  inde- 
pendent of  the  quadrant  in  which  the  angles  lie.  The  student 
should  however,  carry  the  proof  through  for  various  positions  of 
the  terminal  sides  of  the  angles. 

These  addition  formulae  may  now  be  used  in  the  first  place  to 
derive  some  of  the  other  results  of  Chapter  V.  We  have,  for 
instance: 

sin  (a  +  180°)  =  sin  a  cos  180°  +  cos  a  sin  180°  =  -sin  a, 
cos  (a  +  180°)  =  cos  a  cos  180°  -  sin  a  sin  180°  =  -cos  a. 

66.  Subtraction  formulae.  In  the  second  place  we  use  the 
addition  formulae  to  express  the  sine  and  cosine  of  a  —  /3  in  terms 
of  the  sine  and  cosine  of  a  and  /3.  We  know  from  55,  (2)  that  the 
sine  is  an  odd  function  and  that  the  cosine  is  an  even  function,  i.e., 
that 

sin  (  — /3)  =  —sin  /3,  cos  (  —  (3)  =  cos  j3. 

We  replace  /3  by  —  /3  in  the  addition  formulae,  and  make  use  of 
the  above  formulae.     In  this  way  we  find: 

COS  (a  —  8)  =  cos  [a  +  (  —  &)]  =  cos  a  cos  (  —  8)  —  sin  a  sin  (— /3) 

=  cos  a  cos  8  +  sin  a  sin  /S, 

sin  (a  —  8)  =  'si«  [«  +  (  —  8)]  =  s'n  a  ('OS  (  —  8)  +  cos  a  sin  (—  8) 
=  sin  a  cos  8  —  eos  a  sin  8- 

67.  Exercises. 

1.  Express  sin  7.5°  and  cos  75°  in  terms  of  the  ratios  of  4.")°  and  30°,  and 
use  the  results  for  calculating  the  ratios  of  75°. 

2.  Calculate  the  ratios  for  105°. 

3.  Calculate  the  ratios  for  15°.     Compare  the  results  with  those  obtained 
from  the  tables. 

4.  Determine   the  angles   in   the   third   quadrant   whose   ratios   may   be 
calculated  without  the  use  of  tables,  by-means  of  the  addition  and  subtraction 

formulae. 


58  ADDITION  AND   SUBTRACTION   FORMULAE 

Verify  the  following  formulae  by  means  of  the  addition  and  subtraction 
formulae: 

5.  cos  (180°  -  0)  =  -cos  0.  7.   cos  (270°  ±  0)  =  ±  sin  0. 

6.  sin  (180°  -  0)  =  sin  0.  8.   sin  (270°  -  6)  =  -  cos  0. 
9.   sin  (0  +  45°)  +  sin  (0  -  45°)  =  V2  sin  0. 

10.  cos  (9  +  45°)  +  cos  (0  -  45°)  =  V2  cos  0. 

11.  sin  (0  +  30°)  +  cos  (9  +  00°)  =  cos  9. 

12.  sin  (0  -  60°)  =  -cos  {9  +  30°). 

13.  cos  (45°  +9)  =  sin  (45°  -  9)  =  §V2  (cos  9  -  sin  9). 

14.  sin  (45°  +  9)  =  cos  (45°  -  (9)  =  £V2  (cos  9  +  sin  0). 
16.  COS  (a  +  0)  cos  (a  —  Id)  =  COS2  a  -f-  cos-  0  —  1. 

16.  sin  (a  -4-  0)  sin  (a  —  0)  =  sin2  a  —  sin2  /3  =  cos2  0  —  cos2  a 

17.  cos  (nTr  +0)  =  ( -1)"  cos  0.  19.    cos  (nir  -  0)  =  (-1)"  cos0. 

18.  sin  (nw  +  9)  =  (-1)"  sin  0.  20.   sin  (titt  -  0)  =  (-1)"+'  sin  d. 

21.  cos  [(2  n  4- 1)  x/2  -  9]  =  ( - 1)»  sin  0. 

22.  sin  [(2  n  +  1)  tt/2  -  0]  =  ( - 1)»  cos  0. 

23.  cos  [(2ra  +  1)  tt/2  +0]  =  (-1)"+1  sin  0. 

24.  sin  [(2  n  +  1)  tt/2  4-  0]  =  ( - 1)»  cos  0. 

68.  Addition  and  subtraction  formulae  for  the  tangent  and 
cotangent.  Since  tan  6  =  sin  0/cos  9  and  cot  6  =  cos  6 /sin  9 
(Why?),  the  addition  and  subtraction  formulae  for  the  sine  and 
cosine  enable  us  to  obtain  corresponding  formulae  for  the  tangent 
and  cotangent,  viz., 


tan  (a  +  |8) 

sin  (a  +  0) 
cos  (a  4-  /3) 

sin  a  cos  /3  +  cos  a  sin  /3 

cos  a  cos  j8  —  sin  a  sin  13 

and 

cot  (a  4-  j3) 

cos  (a  +  /3) 
sin  (a  +  0) 

cos  a  cos  /3  —  sin  a  sin  /3 
sin  a  cos  /3  4-  cos  a  sin  /3 

The  results  may  be  expressed  in  terms  of  tangents  alone  by 
dividing  the  numerators  and  the  denominators  of  each  of  these 
fractions  by  cos  a  cos  /3,  or  in  terms  of  cotangents  alone  by 
dividing  them  by  sin  a  sin  (3.     In  this  manner  we  find: 

sin  a  cos  0       cos  a.  sin  0 
,      ,      .        cos  a  cos  0       cos  a  cos  0  tan  a  4-  tan  0 

tan  (a  +  0)  = 


cos  a  cos  0   sin  a  sin  0   1  —  tan  a  tan  0 

COS  a  cos  0   cos  a  COS  0 


DOUBLE   AND   HALF-ANGLE   FORMULAE  59 

and 

cos  a  cos  i3       sin  a  tan  0 
.   _  sin  a  sin  ft       sin  a  sin  fl  _  COt  a  cot  0  —  1 

a  —  sin  a  cos  0  .   cos  «  sin  0        cot  a  +  cot  0 

sin  a  sin  0       sin  a  sin  0 

Starting  with  the  formulae 

ON        sin  (a  —  /3)  ,         ,  ,         rtS       cos  (a  —  /S) 

tan  (a  -  0)  =  t ^     and     cot  (a  -  3)  =  ^— r £(' 

cos  (a  —  |3)  sin  (a  —  3) 

and  proceeding  in  exactly  the  same  way  as  above,  we  find: 

-v        tan «  -  tan  /J  .  .        a.      cot  a  cot  0  +  1 

tan  (a  -  /s)  =  r-r-: j — ^    and    cot  («-£)  =  — j— ^-r1 — 

1  -(-  tan  a  tan  3  cot  0  -  cot  a 

69.   Exercises. 

1.  Determine  can  75°  and  cot  75°  by  means  of  the  addition  formulae  for 
the  tangent  and  cotangent. 

2.  Determine  tan  15 '  and  cot  15°  by  means  of  the  subtraction  formulae. 

3.  Determine   tan  105°  and  cot  105°.     (It   would   not  be  advisable  to 
write  90°  +  15°  in  place  of  105°  in  this  problem.     Why  not?) 

,       1  +  tan  9       cot  9  +  1 


4.   Prove  that  tan  (45°  +  9)  =  cot  (45°  -  9) 
6.   Also  that  tan  (45°  -  9)  =  cot  (45°  +  9)  = 


1  -  tan  9       cot  9-1 
1  -  tan  9       cot  9  -  1 


1  +  tan  9      cot  9  +  1 

6.  Express  tan  (a  +  0)  in  terms  of  the  cotangents  of  a  and  0. 

7.  Express  cot  (a  +  0)  in  terms  of  tan  a  and  tan  0. 

8.  Prove  that  tan  (30°  +  6)  =  cot  (60°  -  9)  =  L±_  3  tan  9  . 

v  3  -  tan  6> 

9.  Prove  that  tan  (60°  +  0)  =  cot  (30°  -  6)  =  1+      3cotfl 

cot  0  -  V3 

70.  Double  angle  formulae  and  half-angle  formulae.  From 
the  addition  formulae  we  derive,  by  putting  a  =  @  =  9,  the  fol- 
lowing formulae,  by  means  of  which  our  knowledge  of  the  ratios 
of  any  angle  enables  us  to  find  the  ratios  of  an  angle  twice  as 
large;  in  other  words,  formulae  which  express  the  ratios  of  any 
angle  in  terms  of  the  ratios  of  an  angle  half  as  large;    we  find: 


60  DOUBLE  AND   HALF-ANGLE  FORMULAE 

cos  2  0  =  cos  (0  +  0)  =  cos  0  cos  e  -  sin  0  sin  e  =  cos2  0  -  sin2  e. 

Sin  2  0  =  sin  (0  +  0)  =  sin  0  cos  0  +  cos  0  sin  0  =  2  sin  0  COS  0, 
tan  0  +  tan  0  2  tan  0 

tan  *  °~  1  -  tan  0  tan  0  ~  1  -  tan2  0 
If  we  put  8  =  #/2,  and  therefore  2  0  =  <£,  we  obtain  an  equiva- 
lent form  of  these  formulae,  bringing  out  more  vividly  the  fact 
that  they  express  the  ratios  of  an  arbitrary  .angle  in  terms  of  the 
ratios  of  one  half  of  that  angle,  viz., 

sin  8  =  2  sin  -  cos  ~ ,  cos  8  =  cos2  ~  —  sin2  = ,  tan  8  = > 

22  2  2  i_tan2^ 

where  we  have  again  written  8  in  place  of  <f>. 

The  second  of  these  formulae  leads  to  another  important  set  of 
results,  in  the  following  manner: 

To  the  two  members  of  the  identity 

cos  8  =  cos2-  —  sin2-) 
we  add  the  corresponding  members  of  the  identity 

1  o  0      I  ■      9  ^ 

1  =  cos2^  +  sin--) 

a 

yielding  the  result  1  +  cos  8  =  2  cos2-  (1) 

If  we  subtract  the  first  of  these  identities  from  the  second  we 

obtain:  1  „   .  J  ,0, 

1  —  cos  8  =  2  sin2  j:  •  (2) 

Upon  solving  the  resulting  identities  (1)  and  (2)  for  cos  8/2 
and  sin  8/2  respectively,  we  obtain: 

e  i  /l  +  COS  6  ,      9  ,  4  /l  -  COS  0 

formulae  which  express  the  ratios  of  one  half  of  an  angle  in  terms 
of  the  ratios  of  that  angle.  The  plus  or  minus  sign  is  to  be  used 
according  to  the  quadrant  in  which  8/2  falls. 

Dividing  the  second  of  the  latter  formulae  by  the  first,  we  find: 

VF 


,0         i  /l  —  cos0         sin  0         1  -  cos  0 
tan  -  =  ±* 


2  ~  V  l  +  cos  0  1  +  cos  0  sin  0 
the  last  two  forms  being  derived  from  the  first  by  multiplying 
the  numerator  and  the  denominator  of  the  fraction  under  the 
radical  sign  by  1  +  cos  8  and  1  —  cos  8  respectively.  In  the  last 
two  expressions  the  double  sign  is  not  necessary,  because  1  +-  cos  6 


FACTORIZATION   FORMULAE  61 

and  1  —  cos  0  are  always  positive,  while  tan  0/2  and  sin  0  always 
have  the  same  sign. 

71.   Exercises. 

1.  Determine  the  functions  of  30°  by  means  of  those  of  60°. 

2.  Determine  the  ratios  of  30°  by  means  of  those  of  15°,  found  in  67,  3. 

3.  From  the  ratios  of  15°,  derive  those  of  7°  30'. 

4.  Obtain  a  formula  which  expresses  cot  0/2  rationally  in  terms  of  sin  9 
and  cos  0. 

5.  Prove  that  cosec  19  =  \  sec  9  cosec  9. 


6.  •  Prove  that  sec  x  =  ±  \/  r— r • 

2  »   1  +  sec  d 

7.  Derive  the  double  angle  formula  for  the  tangent  from  the  double 
angle  formulae  for  sine  and  cosine. 

8.  Derive  the  half  angle  formula  for  the  tangent  from  the  double  angle 
formula  for  the  tangent. 

72.  Factorization  formulae.  We  return  once  more  to  the  addi- 
tion formulae  for  the  sine  and  cosine  in  order  to  derive  from  them 
a  set  of  formulae  which  will  enable  us  to  convert  the  sum  (or  the 
difference)  of  the  sines  or  cosines  of  two  angles  into  a  product. 
Such  a  conversion  is  of  great  importance,  as  is  evident  from 
Theorem  I  of  Chapter  III,  whenever  we  wish  to  carry  out  loga- 
rithmic calculations  with  expressions  which  involve  sums  or  differ- 
ences of  sines  or  cosines,  and  also  for  many  theoretical  purposes. 

We  know: 

sin  (a  +  (3)  =  sin  a  cos  /3  +  cos  a  sin  0, 
and  sin  («  —  /?)  =  sin  a  cos  /3  —  cos  a  sin  /3. 

Adding  the  corresponding  members  of  these  two  identities,  we 
find: 

(1)  sin  (a  +  J3)  +  sin  (a  —  (3)  =  2  sin  a  cos  /3. 
Subtracting  them,  we  find: 

(2)  sin  (a  +  /3)  —  sin  (a  —  0)  =  2  cos  a  sin  (3. 

These  formulae  can  be  put  in  a  slightly  different  form,  more 
useful  for  the  purpose  for  which  we  are  deriving  them,  by  putting 


62  EXERCISES 

a  +  ft  =  6  and  a  —  ft  =  <£,  whence  we  obtain  by  addition  and 
subtraction 

e  +  <f>  0-0 

a  =  — ^-^     and     /3  =  — s 


Substitution  of  these  values  for  a  +  /3,  a  —  ft, «  and  /3  in  formulae 
(1)  and  (2)  gives  us  the  following  results: 

sin  6  +  sin  <t>  =  2  sin  ■  "t     cos  ■         » 

and  sin  0  —  sin  <f>  =  2  cos  — 5-^  sin  —5-^  • 

We  now  proceed  in  exactly  the  same  manner  with  the  formulae 

cos  (a  +  ft)  =  cos  a  cos  ft  —  sin  a  sin  /3, 
and  cos  (a  —  ft)  =  cos  a  cos  ft  +  sin  a  sin  ft, 

and  find  the  formulae: 

cos  0  +  cos  0=2  cos  — ~-  cos  ■ 


2  2 

and  cos  0  —  cos  4,  =  —  2  sin    "I     sin     „     » 

73.   Exercises. 

Convert  the  following  sums  and  differences  into  products: 

1.  sin  55°  +  sin  65°.  4.  cos  87°  +  cos  42°. 

2.  cos  75°  -  cos  15°.  5.  cos  312°  -  cos  252°. 

3.  sin  27°  -  sin  18°.  6.  sin  213°  +  sin  237°. 
Calculate  by  means  of  logarithms: 

7.    (sin490  +  sin35°)(cos37°-cos510).  9.  (cos  137°  +  cos  84°)3. 

8    "in  57°  + sin  24° .  ±Q     Vcos  306°  -  cos  246°. 

sin  57°  -  sin  24° 

Prove  the  following  identities: 

« «     sin  0  +  sin  <j>       ,      6  +  <j>        0  —  <t> 
11.    r^— ; ~ — :  =  tan  ■ — -= —  cot  — 5 

0  +  <P  .       0  —  <t> 
12. — - =  —tan — - —  tan — ^ — 


sin 

9  - 

sin 

<t> 

cos 

9  - 

cos 

4> 

cos 

9  + 

cos 

<i> 

sin 

0  + 

sin 

<t> 

cos 

9  + 

cos 

<f> 

cos 

0  - 

COS 

<t> 

13.    — =  tan  — — 


„.       COS0  —  COSc/>  6  +  d> 

14.   - — — . —  =   —tan — t—-' 

sin  d  —  sin  (f>  2 


74. 

1. 
2. 
3. 

4. 


MISCELLANEOUS  EXERCISES 
Summary  of  formulae  proved  in  Chapter  VI. 

cos  (a  +  jS)  =  cos  a  cos  /3  —  sin  a  sin  j3, 
sin  (a  +  /3)  =  sin  a  cos  /3  +  sin  /3  cos  a, 
cos  (a  —  j8)  =  cos  a  cos  /3  +  sin  a  sin  /3, 
sin  (a  —  j8)  =  sin  a  cos  /3  —  sin  /S  cos  a, 
tan  a  +  tan  /? 


tan  (a  +  /3)  = 
tan  (a  —  /3)  = 


1  —  tan  a  tan  /3 ' 
tan  a  —  tan  j8 


1  +  tan  a  tan  /3 ' 

7.  cos  2  0  =  cos2  5  -  sin2  0, 

8.  sin  2  9  =  2  sin  0  cos  0, 

2  tan  0 

9.  tan  2  9 


10. 
11. 

12. 
13. 
14. 
15. 
16. 


1  -  tan20' 
cos  0/2  =  j=V(l  +  cos0)/2, 
sin  0/2  =  ±V(1  -  cos  0)/2, 


+      9       l 
tan  -  =  - 

2  sm  0 


cos  0 


sin  0 


sin  0  +  sin  </>  =  2  sin  - 

sin  0  —  sin  0  =  2  cos 

cos  0  +  cos  0  =  2  cos 

cos0  —  cos^  =  —2  sin 


1  +  cos 

+ 


2 

+ 


cos 


2     Sm_2~ 


+ 


cos 


+ 


63 

see  65. 
see  65. 
see  66. 
see  66. 

see  68. 

see  68. 

see  70. 
see  70. 

see  70. 

see  70. 
see  70. 

see  70. 
see  72. 
see  72. 
see  72. 
see  72. 


75.   Miscellaneous  Exercises  on  Chapter  VI.     (The  more  diffi- 
cult examples  are  marked  with  a  *.) 

1.  Determine  the  ratios  of  165°  (a)  by  means  of  the  addition  formulae: 
(b)  by  means  of  the  half-angle  formulae. 

2.  Prove  that  tan  tt/8  =  V2  -  1. 

3.  Prove  the  identity:  sin  (60°  +  6)  -  cos  (30°  +  0)  =  sin  0. 

a     /-i  i     i   .       (-sm  67°  —  sin  34°)  sin  23° 

4.  Calculate: z-^z ■ 

cos  17° 

Prove  the  following  identities: 

5.  cos  3  0  =  4  cos3  d  —  3  cos  6 

6.  sin  3  0=3  sin  0-4  sin3  6. 
sin2  a  —  sin2  fi 


*7.    tan  (a  +  p)  =    . 

sin  a  cos  a  —  sin  (i  cos  tf 

8.    sin2  3  0  —  sin2  2  6  =  sin  5  5  sin  6. 


Mn     1  +  sin0 

10.    - = — „  =  tan 

1  —  sin  0 


64  MISCELLANEOUS  EXERCISES 

ft  _  „      3  tan  0  -  tan3  0 

9.   tan  3  0  =  — — — i-—  • 

1—3  tan2  0 

(5+0- 

11.  sin  (a  +  0  +  7)  =  sin  a  cos  /3  cos  7  +  sin  /3  cos  a  cos  7  +  sin  7  cos  a 

cos  j3  —  sin  a  sin  /3  sin  7. 

12.  cos  (a  +  /3  +  7)  =  cos  a  cos  /3  cos  7  —  sin  a  sin  /3  cos  7  —  sin  /3  sin  7 

cos  a  —  sin  a  sin  7  cos  /3. 

*13.    tan  a  tan  /3  +  tan  /3  tan  7  +  tan  7  tan  a  =  1,  provided  a  +  0  +  7  =  90°. 

*  .  sin  2  a  cos  a  a 

14.   tr-: 7r-  •  ■=—. =  tan  5  • 

1  +  cos  2  a     1  +  cos  a  2 

16.   cos  (0  +  30°)  +  sin  (6  +  240°)  =  -  sin  6. 

16.  sin  (60°  +6)  +  cos  (6  +  30°)  =  V3  cos  6. 

2 

17.  — — — -r  =  sin  a. 

cot  a/2  -j-  tan  a/ 2 

18.  sin  0  +  sin  (0  +  2  jr/3)  +  sin  (0  +  4  tt/3)  =  0. 

19.  cos  0  +  cos  (0  +  2  tt/3)  +  cos  (0  +  4  tt/3)  =  0. 

20.  sin  4  0  =  4  sin  0  cos3  0  —  4  sin3  0  cos  0. 

21.  cos  (a+/?)+  cos  (a  —  /3)  +  cos  (—  a  + /3)+  cos  (—  a  —  0)  =  4  COS  a  cos  0. 

„„      1  -  tan2  a/2 

22.  -—— — - — --  =  cos  a. 
1  +  tan2  a/2 

23.  tan  (0  +  45°)  +  tan  (0  -  45°)  =  2  tan  2  0. 

*24.    tan  2  a  —  tan  a  = ; s —  • 

cos  a  +  cos  6  a 

25.  cos  4  0  =  8  cos4  0-8  cos2  0  +  1. 

*26.  sin  2  a  +  sin  2  j3  +  sin  2  7  =  4  sin  a  sin  0  sin  7,  if  a  +  0  +  7  =  180°. 

27.  sin  3  0  =  4  sin  0  sin  (tt/3  +  0)  sin  (tt/3  -  0). 

28.  Calculate  the  functions  of  3  w,  8. 

*29.    tan  (tt/4  +  0/2)  +  cot  (V/4  +  0/2)  =  2  sec  0. 

*30.    cos  2  a  +  cos  2  (3  +  cos  2  7  =   —  1  —  4  cos  a  cos  /3  cos  7,  if  a  +  /3  + 
7  =  180°. 


CHAPTER  VII 

THE  SOLUTION   OF  TRIANGLES 

76.  The  Law  of  Sines;  the  area  of  a  triangle.  We  consider 
in  this  chapter  the  problem  of  "  solving  an  arbitrary  triangle 
ABC,"  i.e.,  the  determination  of  the  unknown  elements  of  a  tri- 
angle ABC  of  which  sufficient  elements  are  given.  The  problem 
is  solved  by  the  use  of  various  relations  subsisting  between  the 
sides  and  angles  of  a  triangle. 


*~x 


Fig.  38a 

Denoting  the  length  of  the  perpendicular  dropped  from  the 
^vertex  A  upon  the  opposite  side  BC  by  ha,  and  using  analogous 
notations  for  the  other  perpendiculars,  we  have, 
(1)         ha  =  c  sin  B  =  b  sin  C,         hb  =  a  sin  C  =  c  sin  A, 
hc  =  b  sin  A  =  a  sin  B. 
For  ha  in  Fig.  38a,  the  above  result  is  obtained  most  readily,  if 
we  remember  that   Z  B  is  placed  in  standard  position  with  refer- 
ence to  the  axes  indicated  in  the  diagram. 
From  these  formulae  we  derive: 

(a)  Theorem  I.  The  sines  of  the  angles  of  a  triangle  are  propor- 
tional to  the  sides  opposite  the  angles.*     (Law  of  Sinks.) 

*  It  is  a  familiar  theorem  of  plane  geometry  that  of  two  unequal  sides  of 
a  triangle  the  greater  side  lies  opposite  the  greater  angle.  The  law  of  sines 
may  be  looked  upon  as  completing  this  theorem  by  stating  how  the  unequal 
sides  are  related  to  the  angles  opposite  them. 


66  TWO   ANGLES  AND  ONE  SIDE 

To  prove  this  law,  we  divide  the  two  expressions  for  ha  in  equa- 
tions (1)  by  be,  those  for  hb  by  ac.     In  this  way  we  obtain: 

sin  B/b  =  sin  C/c  =  sin  A  /a, 

which  was  to  be  proved. 

(b)  Theorem  II.  The  area  of  a  triangle  is  equal  to  one  half  of  the 
product  of  any  two  sides  multiplied  by  the  sine  of  the  angle  included 
by  them. 

Proof.     Denoting  the  area  of  triangle  ABC  by  A,  we  have: 

A  =  a/2  •  ha  =  6/2  •  hb  =  c/2  •  hc. 

If  in  these  expressions  we  substitute  for  the  altitudes  ha,  hb  and 
hc  the  values  which  they  have  in  equations  (1),  we  find 

A  =  \  ab  sin  C  =  \  be  sin  A  =  \  ca  sin  B, 

which  was  to  be  proved. 

77.  Two  angles  and  one  side.  A  triangle  is  determined  when 
two  angles  and  a  side  are  given.  The  law  of  sines  suffices  to  deter- 
mine the  remaining  elements  in  this  case.  For,  suppose  that  Ay 
B,  and  a  are  given;  we  have  then  C  =  180°  —  (.4  +  B). 

Moreover  sin  B/b  =  sin  A/a  and  sin  C/c  =  sin  A  /a;  hence 

,        a  sin  B  ,  a  sin  C 

b  =  — : — -7-  ,     and     c  —  —. — 7- » 
sin  A  smi 

which  completes  the  solution. 

If  the  given  elements  are  measured  by  short  numbers  the  cal- 
culation may  be  carried  out  directly  by  means  of  a  table  of  natural 
values;  the  expressions  for  b  and  c,  however,  are  well  adapted  to 
calculation  by  means  of  logarithms. 

Example. 

Given,     a  =  43.257,    A  -  57°  23',    C  =  49°  47'. 

Required.     B,  b  and  c. 

Solution.     B  =  180°  -  (A  +  C)  =  180°  -  107°  10'  =  72°  50'. 

b  a  .  _  a  sinB  _  43.257    sin  72°  50' 

sin  B  ~  sin  A  '     1,e''        ~    sin  A    '  sin  57°  23' 


a 
sin  A  ' 


EXERCISES  67 

asinC      43.257    sin  49°  47' 


i.e.,    c  = 


sin  C 

log  43.257=   1.63606 
log  sin  72°  50'=   9.98021-10 


sin  A  sin  57°  23' 

log  43.257=   1.63606 
log  sin  49°  47'=   9.88287-10 


11.61627-10 
log  sin  57°  23' =   9.92546-10 


S 


11.51893-10 
log  sin  57°  23'=   9.92546-10 


78. 

1. 

2. 
3. 
4. 
5. 


log  6=   1.69081 
6  =  49.069 


Exercises. 

A  =  39°  27',  B 


logc  =    1.59347 
c  =39.216 


108°  51',  b  =  .43215.     Determine  C,  a,  c. 
C  =  30°,  A  =  45°,  c  =  123.     Determine  B,  a,  b. 
B  =  27°  45'  15",  C  =  89°  19'  20",  c  =■■  14.302.     Determine  A,  a,  b. 
A  =  37°  12'  30",  C  =  58°  26'  40",  a  =  103.47.     Determine  B,  b,  c. 
We  wish  to  determine  the  distance  from  a  point  A  to  a  point  B,  situ- 
ated in  a  marsh,  visible  but  not  accessible  from  A  (see  Fig.  39).     For  this 


purpose,  we  measure  the  distance  from  A  to  a  point  C,  from  which  both  A 
and  B  are  visible  and  we  measure  the  angles  BAC  and  ACB.  Calculate  AB, 
if  AC  =  750  feet,  ABAC  =  32°  27'  and  ZACB  =  47°  12'. 

6.    Devise  a  method  for  finding  the  distance  from  a  point  A  on  one  bank 
of  a  river  to  a  point  B  on  the  opposite  bank. 

T 


Fig.  40 

7.  To  determine  the  height  of  a  tower  7'>S^  (see  Fig.  40)  we  measure  the 
angle  of  elevation  of  its  top  T  from  two  points  A  and  B  lying  on  the  same 
side  of  T,  in  the  same  vertical  plane  with  T  and  a  known  distance  apart. 


68      TWO   SIDES   AND   AN   ANGLE  OPPOSITE  ONE  OF  THEM 

Calculate  the  height  of  the  tower,  if  we  find  that  AB  =  257  feet,  Z  TAB  = 
19°  43'  and  Z  TBS  =  47°  29'.  (Compare  the  present  method  of  solving 
this  problem  with  the  method  used  for  solving  problem  10  in  49.) 

8.    Solve  problem  10  in  49  by  the  method  used  in  the  preceding  problem. 

79.   Two  sides  and  an  angle  opposite  one  of  them.     If  two 

sides  of  a  triangle  and  an  angle  opposite  one  of  them  are  given, 
the  triangle  is  not  completely  determined.  Let  there  be  given 
the  sides  a  and  b  and  the  acute  angle  A  (see  Fig.  41).     To  con- 


struct the  triangle  ABC  we  lay  off  a  line  equal  to  b  on  one  of  the 
legs  of  the  angle  A.  The  vertex  C  is  then  located.  We  then 
strike  an  arc  with  C  as  a  center  and  a  as  a  radius.  The  intersec- 
tion of  this  arc  m  with  the  second  leg  AX  of  angle  A  determines 
the  third  vertex  of  the  triangle,  B. 

It  is  now  clear  that  if  a  is  shorter  than  the  perpendicular  dis- 
tance p  from  C  to  AX,  as  in  Figure  42a,  then  the  arc  m  will  not 


Pig.  42a 


Fig.  42b 


intersect  AX  at  all,  so  that  no  triangle  can  be  constructed.  If  a 
is  equal  to  the  perpendicular  p,  as  in  Figure  42b,  we  obtain  a  single 
right-angled  triangle.  If  a  is  greater  than  p,  but  less  than  b  (see 
Fig.  42c),  then  the  arc  m  will  cut  AX  in  two  points,  B  and  B' , 


TWO  SIDES  AND  AN  ANGLE  OPPOSITE  ONE  OF  THEM       69 

and  we  will  obtain  two  triangles,  ACB  and  ACB' ',  both  of  which 
will  satisfy  the  requirements  of  the  problem.  Since  CB  =  CB', 
Z  CB'B  =  Z  CBB',  and  therefore  the  angles  B  and  B'  occurring 
in  triangles  ABC  and  AB'C  respectively  are  supplementary  angles. 
If  a  >  b  (see  Fig.  42d)  the  arc  m  will  meet  AX  again  at  two  points 


Fig.  42d 


Fig.  42e 


B  and  B' ',  but  one  of  these  points  will  fall  on  XA  produced,  so 
that  the  triangle  AB'C,  to  which  it  gives  rise,  does  not  contain 
the  given  angle  A.  Hence  in  this  case  there  is  only  one  triangle 
Avhich  satisfies  the  requirements  of  the  problem,  viz.,  triangle  ABC. 
If  the  given  angle  A  is  obtuse  the  construction  of  triangle  ABC 
proceeds  in  the  same  way  as  before  (see  Fig.  42e).  It  is  evident 
that  in  this  case  there  is  no  solution  possible,  unless  a>  b.  We 
observe  moreover  that  in  all  the  cases  which  we  have  considered 
the  perpendicular  p  is  equal  to  b  sin  A.  Hence  we  can  put  the 
results  of  this  discussion  in  the  following  form: 

Theorem  III.  If  two  lines  and  an  angle,  such  as  a,  b,  A  are  given, 
then  there  may  be  no  triangle,  one  triangle  or  two  triangles  of  which 
the  given  lines  are  sides  and  of  which  the  given  angle  is  an  angle 
opposite  one  of  these  sides,  viz., 

If  a  <  b  sin  A,  there  is  no  triangle; 

if  a  =  b  sin  A  and  A  is  acute,  there  is  one  triangle; 

if  a  >  b  sin  A,  a  <  b  and  A  is  acute,  there  are  two  triangles; 

if  a  =  b  and  A  Is  acute,  there  is  one  triangle; 

if  «  >  b,  there  is  one  triangle. 


The  different  cases  mentioned  in  this  theorem  are  illustrated 
in  the  examples  which  follow.  The  solution  proceeds  in  each  case 
according  to  the  following  plan: 


70      TWO  SIDES  AND   AN   ANGLE  OPPOSITE  ONE  OF  THEM 

From  the  law  of  sines,  we  find:  sin.  B  = ,  which  enables 

a 

us  to  determine  B.     Next  we  can  calculate  C,  since  C  =  180°  — 

(A  +  B) .    Finally  we  use  the  law  of  sines  to  find  c,  viz.,  c  =  — — j-  • 

sin  A 

If  a  <  b  sin  A,  we  find  that  sin  B  >  1;  hence  no  angle  B  can 
be  found,  and  therefore  no  triangle  exists.  If  a  =  b  sin  A,  then 
sin  B  —  1,  B  =  90°  and  we  have  a  single  right-angled  triangle. 

If  a  >  6  sin  A,  then  sin  B  <  1  and  we  can  determine  angle  B 
from  the  tables.  But,  besides  the  acute  angle  B  found  from  the 
tables,  we  must  also  consider  the  supplement  of  B,  whose  sine  is 
equal  to  the  sine  of  B;  we  have  therefore  B'  —  180°  —  B.  More- 
over if  a  <  b,  then  we  must  have  B  >  A  (see  footnote  on  p.  65); 
hence  both  angles,  B  and  B',  can  be  used,  if  A  is  acute,  while 
neither  can  be  used  if  A  is  right  or  obtuse.  If,  on  the  other  hand, 
a  >  b,  then  we  must  have  B  <  A,  so  that  only  the  acute  angle  B, 
found  from  the  tables,  can  be  used. 

Example  1. 

Given,     a  -  4.73,  b  =  18.65,  A  =  43°  27'. 

Required,     c,  B,  C. 

Solution . 

sin  B       sin  A      xU      ,  .     D       b  sin  A       18.65  sin  43°  27' 

— j —  =  ,     therefore     sin  B  =  =  -7-== > 

6  a  a  4.73 

log  18.65  =     1.27068. 

log  sin  43°  27'  =    9.83741  -  10. 

A 

log  b  sin  A  =  11.10809  -  10. 

log  a  =  log  4.73  =       .67486. 

We  notice  that  log  b  sin  A  >  log  a;  therefore  n  <  b  sin  A. 
Hence  no  triangle  can  be  found  with  the  given  elements. 

Example  2. 

Given,     a  =  14.73,  b  =  18.65,  A  =  43°  27'. 

Required,     c,  B,  C. 


TWO  SIDES   AND   AN  ANGLE  OPPOSITE  ONE  OF  THEM     71 

Solution.     We  have  now 

.     D       6  sin  A       18.65   sin    43°   27' 

sin  B  = = rj-=^ » 

a  14.73 

log  18.65  =     1.27068, 

log  sin  43°  27'  =    9.83741  -  10, 

A 

log  b  sin  A  =  11.10809  -  10, 

log  14.73  -     1.16820, 

S 

log  sin  B  =    9.93989  -  10. 

Here  a  >  b  sin  A,  a  <  b  and  A  is  acute.     Hence  there  are  two 
triangles  in  this  case.     We  find: 

First  solution  Second  solution 

B  =  60°  32'  43".  B'  =  119°  27'  17". 

We  know  A  =  43°  27'.  We  know  A  =  43°  27'. 

Hence        C  =  180° -(A  +  B)      Hence       C"  =  180°  -  (A  +  B) 
=  76°  0'  17".  -  17°  5'  43". 

Furthermore :  Furthermore : 

sin  C      sin  A  .  A,       .  sin  C"       sin  A  .  ,.       , 

-  = }    and  thererore  — , —  =  >  and  therefore 

c  a  c  a 

_ osin  C _  14.73 X sin  76°  17/r      , _ a  sin  C _  14.73 X sin  17°  5'  43" 


sin  A            sin  43°  27'  sin  A               sin  43°  27' 

log  14.73  =  1.16820  log  14.73  =  1.16820 

log  sin  76°  17"  =  9.98691  -  10  log  sin  17°  5' 43"=    9.46829-10 

A  —A 

11.15511  -  10  10.63049  -10 

log  sin  43°  27'  =    9.8374 1  -  10  log  sin  43°  27'   =   9.8374 1-10 

S  s* 

logc=     1.31770  logc'=      .79908 

c  =  20.783  c'=   6.2963 


72     TWO  SIDES  AND  AN   ANGLE  OPPOSITE  ONE  OF  THEM 
Figure  43  is  a  drawing  to  scale  of  the  triangles  ABC  and  AB'C. 


Example  3. 

Given,     a  =  24.73,  b  =  18.65,  A  =  143°  27'. 

Required,     c,  B,  C. 

„  ,    .  .    „       b  sin  A       18.65  sin  143°  27' 

solution,     sin  i)  =  = nA  no 

a  24.73 

log  18.65  =     1.27068 

log  sin  143°  27'  =  log  sin  36°  33'  =    9.77490  -  10 

A 

log  b  sin  A  =  11.04558  -  10 
log  24.73  =     1.39322 

S 

log  sin  B  =    9.65236  -  10 
B  =  26°  41'  14" 

Since  a  >  b  and  A  is  obtuse,  there  is  only  one  solution  and  B 
must  be  acute.  Hence  the  value  found  from  the  table  is  the 
only  one  that  can  be  used  in  this  case. 

C  =  180°  -  (143°  27'  +  26°  41'  14")  =  9°  51'  46" 
asinC      24.73  sin  9°  51' 46" 


c  = 


sin  A  sin  143°  27' 


log  24.73  =     1.39322 
log  sin  9°  51'  46"  =    9.23373  -  10 

A 

log  a  sin  C  =  10.62695  -  10 
log  sin  143°  27'  =    9.77490  -  10 

S 

log  c  =       .85205 
c=    7.1130 


THE  LAW   OF  COSINES 


73 


80.   Exercises. 

1.  a  =  42.3,  b  =  57.03,  A  =  35°  35'.     Determine  c,  B,  C. 

2.  c  =  507.8,  b  =  751.3,  B  =  23°  47'.     Determine  a,  A,  C. 

3.  b  =  5.5,  c  =  4.3,  C  =  75°  29'.     Determine  a,  A,  B. 

4.  a  =  3.207,  c  =  7.831,  C  =  137°  18'.     Determine  b,  A,  B. 

5.  a  =  37.052,  6  =  49.312,  .4  =  19°  25'.     Determine  c,  B,  C. 

6.  c  =  .047031,  b  =  .047031,  C  =  28°  31'.     Determine  a,  A,  B. 

7.  An  inland,  known  to  be  75  miles  wide,  subtends  an  angle  of  40°  17' 
from  a  point  P,  40  miles  distant  from  one  extremity  of  the  island.  How  far 
is  P  from  the  other  extremity  of  the  island? 

8.  A  flagpole,  10  feet  high,  subtends  an  angle  of  2°  37'  from  a  point  A. 
If  A  is  200  feet  from  the  foot  of  the  pole,  how  far  is  it  from  the  top? 

9.  Two  lighthouses,  M  and  L  are  40  miles  apart.  At  8.30  a.m.  a  ship  S 
leaves  M  and  travels  at  the  rate  of  12  miles  per  hour.  At  11  a.m.  the  distance 
between  M  and  L  subtends  an  angle  of  33°  from  the  ship.  How  far  is  S 
from  L  at  that  moment? 


81.  The  Law  of  Cosines.  Two  sides  and  the  included  angle. 
Three  sides.  If  three  sides  of  a  triangle  are  given,  and  also  if 
two  sides  and  the  included  angle  are  given,  the  triangle  is  deter- 
mined. The  law  of  sines  does  not  suffice,  however,  to  calculate 
the  unknown  elements  in  these  cases.  We  therefore  proceed  to 
derive  a  new  relation  between  the  sides  and  angles  of  the  triangle. 


Fig.  44a 

In  Fig.  44a,  the  obtuse  angle  B  is  in  standard  position  with 
reference  to  the  axes  indicated  in  the  diagram.  We  have,  there- 
fore, in  both  figures: 

BD  =  c  cos  B,  ha  =  c  sin  B,  BC  =  a. 

Using  now  Theorem  I  of  5.  we  find  that  BD  +  DC  +  CB  =  0. 
Therefore,  DC  =  -  BD  -  CB  =  BC  -  BD  =  a  -  c  cos  B.  (1) 
Moreover  b2  =  ha2  +  W"1  =  c2  sin2  B  +  L>C2.  (2) 


74  THE  LAW  OF   COSINES 

Substituting  (1)  in  (2),  we  find: 

b2  =  c2  sin2  B  +  (c  cos  B  —  a)2 

=  c2  sin2  B  -\-  c2  cos2  5  +  a2  —  2  ac  cos  5; 
i.e.,  b2  =  e2  +  a2  -  2  oc  cos  B, 

We  have  therefore  proved  the  following  theorem: 

Theorem  IV.  The  square  of  one  side  of  a  triangle  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides  diminished  by  twice  the 
continued  product  of  these  two  sides  and  the  cosine  of  the  angle 
included  by  them.     (Law  op  Cosines.) 

By  dropping  perpendiculars  from  each  of  the  other  two  vertices 
of  triangle  ABC  and  proceeding  in  an  exactly  similar  manner,  we 
obtain  two  formulae  analogous  to  the  one  derived  above,  viz., 

c2  =  a2  -f  b2  —  2  ab  cos  C        and       a2  —  b2  +  c2  —  2  be  cos  A. 

In  the  form  in  which  they  are  here  given,  these  formulae  enable 
us  to  calculate  the  third  side  of  a  triangle  of  which  two  sides  and 
the  included  angle  are  known.  If  we  solve  them  for  the  cosines 
of  the  angles,  we  obtain: 

.     tf+c2— a2  „     c2+o2  — />2  „     a2-\-b2  —  c2       ._. 

cos  A  =  — — ; ,    cos  B  =  — - ,     cos  C= — ^—. »    (3) 

2  be  2  ca  2  ab 

in  which  form  they  are  well  adapted  to  determining  the  angles  of  a 
triangle  whose  sides  are  given. 

We  observe  that  none  of  the  formulae  derived  in  the  present 
section  are  suited  to  logarithmic  calculation.  For  this  reason 
they  are  useful  only  for  computations  which  involve  short  numbers. 

Example  1. 

Given,     a  =  14,  b  =  27,  C  =  35°. 

Required,     c,  A,  B. 

Solution.     By  means  of  Theorem  IV  we  find: 

C2  =  a2  +  ?,2  _  2  ab  cos  C  =  14-  +  272  -  2  •  14  •  27  •  cos  35°; 

C2  =  196  +  729  -  756  X  .81915  =  925  -  619.27740  =  305.72260. 

c   =  V  305.72260  =  17.484. 


THE  LAW  OF  COSINES  75 

To  complete  the  calculation,  we  use  the  law  of  sines,  and  find: 

.       a  sin  C  ,  .     „       b  sin  C 

sin  A  = j  and         sin  B  =  — • 

c  c 

log  14  =     1.14613  log  27  =     1.43136 

log  sin  35°  =    9.75859  -  10  log  sin  35°  =    9.75859  -  10 

-A  -A 

S 


10.90472  -  10  11.18995  -  10 

log  17.484  =     1.24264  log  17.484  =     1.24264 


log  sin  A  =    9.66208  -  10  log  sin  B  =    9.94731  -  10 

A  =  27°  20'  28"  or  152°  39'  32"  and  B  =  62°  20'  34"  or  1 17°  39'  26". 

Since  a  <  c,  we  must  have  A  <  C;  hence  A  =  27°  20'  28". 

Since  b  >  a  and  b  >  c,  we  must  have  B  >  A  and  B  >  C;  both 
values  of  B  satisfy  this  condition;  it  is  clear,  however,  that  with 
the  smaller  value  of  B,  the  condition  that  A  +  B  +  C  =  180° 
would  not  be  satisfied;   therefore  B  =  117°  39'  26". 

Example  2. 

Given,     a  =  10,  b  =  15,  c  =  19. 

Required.     A,  B,  C. 

Solution.     Using  formula  (3),  we  find: 

r      a2  +  62  -  &      100  +  225  -  361       -36  tonnn 

cosC  =  -2flr  ^ =  m=-J20oa 

Since  cos  C  is  negative,  Z  C  is  obtuse  and  cos  (180°  —  C)  = 
-cos  C  =  .12000. 
We  find  from  the  tables,  that  180°  -  C  =  83°  6'  29";  and  hence 
C  =  96°  53'  31". 
Furthermore, 

„       c2  +  a2  -  b2      361  +  100  -  225      236       __ 


2  ca 

380 

380 

.UiilUfJ, 

therefore 

B  =  51°  36'  27". 

And  finally, 

h1    +    (.2_ 

a2      225  +  361  - 

100 

486 
~  570  " 

.85263; 

(°"-1               2  be 

570 

whence  we  obtain  A  -  31°  30'  4". 

Check.     A  +  B  +  C  =  180°  0'  2". 


76  EXERCISES 

82.   Exercises. 

1.  a  =  120,  b  =  150,  C  =  60°.     Determine  c,  A,  B. 

2.  p  =  1.3,  q  =  1.4,  r  =  1.5.     Determine  P,  Q,  R. 

3.  o  =  17,  6  =  15,  c  =  29.     Determine  A,  B,  C. 

4.  r   =  .45,  s  =  .78,  T  =  45°.     Determine  t,  R,  S. 

5.  To  determine  the  width  of  a  lake,  the  distances  of  its  extreme  points 
A  and  B  from  a  point  P  and  the  angle  subtended  by  AB  at  P  are  measured. 
It  is  found  that  AP  =  750  feet,  BP  =  600  feet,  and  LP  =  32°. 

6.  It  is  desired  to  make  a  triangle  out  of  sticks  that  are  5,  8,  and  9  inches 
long.  What  angle  should  the  first  two  of  these  sticks  make,  in  order  that 
the  third  one  may  be  just  long  enough  to  join  their  free  ends? 

7.  Prove  that  a-  =  b2  +  c2  —  2  be  cos  A. 

8.  Prove  that  a2  +  b2  +  c2  =  2  ab  cos  C  +  2  be  cos  A  +  2  ca  cos  B. 

a2  +  b2  -  c2 


9.    Prove  that  cos  C  = 


2  ab 


83.  Summary  and  critique.  In  the  preceding  sections  we  have 
learned  to  calculate  the  unknown  parts  of  triangles  of  which  are 
given  (a)  one  side  and  two  angles  (77);  (6)  two  sides  and  an 
angle  opposite  one  of  the  sides  (79) ;  (c)  two  sides  and  the  included 
angle  (81) ;  or  (d)  three  sides  (81) .  We  know,  moreover,  from  the 
study  of  plane  geometry,  that  if  a  triangle  is  to  be  determined 
by  sides  and  angles,  then  the  given  elements  must  form  one  of 
the  four  sets  (a),  (6),  (c)  or  (d)  enumerated  above.  Hence  the 
general  problem  of  "solving  a  triangle"  has  been  solved  in  sec- 
tions 77-81,  in  so  far  as  it  relates  to  triangles  determined  by 
means  of  sides  and  angles.  There  arc1  however  two  criticisms  to 
be  made  of  the  theory  developed  so  far,  viz. : 

(1)  The  methods  developed  in  81  for  cases  (c)  and  (d)  are  not 
suited  to  the  use  of  logarithms  and  are  not  very  useful,  therefore, 
in  problems  involving  long  numbers. 

(2)  There  are  no  convenient  methods  for  accurately  checking 
the  calculations  in  cases  (a),  (b)  and  (c). 

In  order  to  meet  these  criticisms  some  further  relations  between 
the  sides  and  angles  of  a  triangle  will  now  be  derived. 

84.  The  law  of  tangents.     From  Theorem  I,  we  conclude  that 

sin  A  _  a 

siiT#  ~  b  '  (  j 


THE  LAW  OF  TANGENTS  77 

Adding  1  to  both  sides  of  this  equation,  we  obtain: 


sin  A  -f  sin  B  _  a  -f-  b 
sin  B  ~b~' 

subtracting  1  from  both  sides  of  equation  (1)  gives  us: 

sin  A  —  sin  B  _  a  —  b 
sin  B  b 


(2) 


(3) 


If  we  divide  the  sides  of  (2)  by  the  corresponding  sides  of  (3),  we 
find: 

sin  A  +  sin  B      a  +  b 


sin  A  —  sin  B      a  —  b 


(4) 


We  now  make  use  of  formulae  (13)  and  (14)  of  74  to  factor 
respectively  the  numerator  and  denominator  on  the  left  hand 
side  of  (4);  we  also  divide  the  numerator  and  denominator  by  2 
and  obtain: 

a  +  b  _  sin  \  (A  +  B)  cos  \  (A  -  B) 

a  -  b  ~  sin  \  (A  -  B)  cos  \  (A  +  B) 

tan  %(A+B)  cot  \  C 


tan  \{A  -B)      tan  |(A  -  B) 
To  justify  the  last  step  we  observe  that  A  -J-  B  +  C  =  180°. 

Hence 

\{A  +  B)  =  90°  -  \  C.  (5) 

.-.     tan  |  (A  +  B)  =  tan  (90°  -  \  C)  =  cot  \  C  (see  60,  (4)). 
The  result  is  most  conveniently  written  in  the  following  form, 

tan  |  {A  -  R)  =  ^-£  cot  *  C' 

a  -\-  b 

which  formula  is  expressed  in  the  following  theorem: 

Theokkm  V.  The  tangent  of  one  half  the  difference  of  two  angles 
of  a  triangle  is  equal  to  the  quotient  of  the  difference  of  the  sides 
opposite  these  angles  by  their  sum,  multiplied  by  the  cotangent  of 
one  half  the  angle  included  by  them.     (Law  of  Tangents.) 


78  MOLLWEIDE'S  EQUATIONS 

85.   Mollweide's  Equations.    If  we  multiply  both  sides  of  equa- 
tions (2)  and  (3)  of  84  by  the  corresponding  sides  of  the  equation 

sin  B      b  c    ,    , 

—. — 7=  =  - ,  we  nnd  that 
sin  C      c 

sin  A  +  sin  B  _  a  -f-  b 

sin  C  c 

and 

sin  A  —  sin  B      a  —  b 


sin  C 


(1) 

(2) 


We  factor  the  numerators  on  the  left  hand  sides  of  these  equa- 
tions by  means  of  Formulae  (13)  and  (14)  of  74.  The  denomina- 
tors we  change  by  writing  sin  C  =  2  sin  \  C  cos  \  C,  which  is  a 
consequence  of  formula  (8)  of  74.  In  this  way  equations  (1)  and 
(2)  will  assume  the  following  form: 


(3) 
(4) 


sin  j  (A  +  B)  cos  \  (A  -  B)  _  a  +  b 

sin  \  C  cos  \  C  c 

and 

sin  %  (A  -  B)  cos  %  (A  +  B)      a  -b 

sin  \  C  cos  |  C  c 

Moreover,  from  equation  (5)  of  84,  it  follows  that 
sin  i  (A  +  B)  =  cos  \  C 
and  cos  \  (A  +  B)  =  sin  \  C,  (see  19  and  55). 

Consequently,  equations  (3)  and  (4)  may  be  simplified  to  the 
final  form: 

a  +  b  _  cos  \  (A  -  B)  a  -  b  _  sin  \  (A  -  li) 

—         —  -.     z  7;      ■    and    —  — ^ — 77 —  • 

c  sin  §  C  c  cos  §  C 

These  equations  are  known  as  Mollweide's  equations. 

86.  Exercises. 

1.   Prove:  tan  \  (C-A)  =  C-^--  cot  \  B. 

c  +  a 

cos  -J  yl  a 

3.    Derive  a  proof  of  the  law  of  tangents  from  Mollweide's  equations. 


TWO  SIDES  AND  THE  INCLUDED   ANGLE  79 

87.  Two  sides  and  the  included  angle.  The  law  of  tangents 
enables  us  to  meet  the  criticism  brought  forward  in  83  with  refer- 
ence to  case  (c).  Moreover,  Mollweide's  equations  are  well 
adapted  to  serve  for  checking  the  calculations  in  cases  (a),  (6) 
and  (c),  because  they  involve  all  the  sides  and  all  the  angles  of 
the  triangle. 

Example. 

Given,     a  =  .4503,  b  =  .7831,  C  =  43°  48'. 
Required,     c,  A,  B. 

Solution.  We  use  the  law  of  tangents  to  determine  angles  A 
and  B: 

tan  i(p-A)-*=£  cot  |  C  =  jfH  COt  21°  34'- 
log  .3328  -  9.52218-10 
log  1.2334  =    .09110 

S 

9.43108-10 
log  cot  21°  54'  =    .39578 

A 

log  tan  \  (B  -  A)  =  9.82686-10 
\  (B  -  A)  =33°  52'  11", 
but, 

\  (£+A)  =  90°-i  C  =  90°-21°  54'  =  68°  6' 

A  and  S. 

Therefore, 

Z  B  =  h(B  -i-  A)  +  |  (£  -  A)  =  101°  58'  11" 
and 

Zi  =  f(5  +  i)-|(B-4)=    34°  13'  49". 

Furthermore: 

_  a  sin  C  _  .4503  X  sin  43°  48r 

sin  A    ~     sin  34°  13'  49" 

log  .4503  =  9.65350-10 

log  sin  43°  48'  =  9.84020-10 
A 


19.49370-20 
log  sin  34°  13'  49"  =  9.75014-10 


logc  =  9.74356-10 
c  =    .5.5406 


80  THE  HALF-ANGLE  FORMULAE 

Ckeck.  aM(B -C)=L=i. 

cos  f  A  a 

\{B-C)  =  29°  5'  6",         I A  =  17°  6'  55",  6  -  c  =  .22904 

log  sin  H#-C)    =  9.68673  -  10         log  (6  -  c)  =  9.35992  -  10 

log  cos  §  A  =  9.98032  -  10  log  a  =  9.65350  -  10 

S S 


9.70641  -  10  9.70642  -  10 

88.  Exercises. 

1.  c  =  27.04,  b  =  84.31,  A  =  112°  44'.     Determine  a,  B,  C. 

2.  a  =  3152,  c  =  4281,  B  =  88°  27'.     Determine  6,  A,  C. 

3.  p  =  .0432,  <?  =  .0586,  R  =  47°  36'.     Determine  r,  P,  Q. 

4.  x  =  .8132,  z  =  .5817,  F  =  120°.     Determine  y,  X,  Z. 

6.  A  hill  slopes  at  an  angle  of  17°.  A  point  P  is  39.3  feet  up  the  hillside; 
a  point  Q  is  in  the  plane,  173.5  feet  from  the  foot  of  the  hill  and  in  the  same 
vertical  plane  as  P.     How  far  is  P  from  Q? 

6.  The  distance  from  the  earth  to  the  sun  is  approximately  92.9  million 
miles,  the  distance  from  the  earth  to  the  moon  approximately  239,000  miles. 
What  is  the  distance  from  the  sun  to  the  moon  at  a  moment  when  the  line 
sun-moon  subtends  at  the  earth  an  angle  of  24°  36'? 

89.  The  half-angle  formulae  for  the  angles  of  a  triangle.     We 

start  with  the  first  formula  for  tan  \  A  given  in  70  and  multiply 
numerator  and  denominator  of  the  fraction  under  the  radical 
sign  by  2  be.     In  this  way  we  find : 


,    ,        .  /l  —  cos  A       .  /2  be  —  2  be  cos  A  /1N 

tan  \  A  =  \/rVc^A  =  \/2bc  +  2bc  cosA'  (D 

If  we  multiply  both  sides  of  formula  (3)  of  81  by  2  be,  we  obtain: 

2  be  cos  A  =  b2  +  c2  -  a1.  (2) 

We  substitute  (2)  in  (1);  this  gives  us: 

i  /2  bc  +  a2  -  b2  -  c2  =     /a2  -  (ft2  -~2~bc  +  c2) 

tan  2  A  ----  y  2  bc  +  f)2  +  c,  _ -.,      y  2  5c  +       _  fl2 


v/ 


(q  +  6  -  c)  (a  -  b  +  c) 

(6  +  c  +  a)  (6  +  c  -a)'  W 


THREE  SIDES  81 

Now,  we  introduce  the  abbreviation  2  s  for  the  perimeter  of 
the  triangle;    i.e.,  a  +  b  +  c  =  2  s.     Subtracting  in  turn  2  a,  2  b 
and  2  c  from  both  sides  of  this  equality,  we  obtain : 
—  a-\-b-{-c  =  2  (s  —  a), 
a  -  6  +  c  =  2  (s  -  6), 
and  a  4-  6  —  c  =  2  (s  —  c). 

These  expressions  are  now  substituted  in  equation  (3),  which 
then  becomes: 

»         s  (s  —  a)  s  —  a       »  s  *  —  a 


where   P  =  \/(s  ~  q)  (s  ~  b)  {s  ~  c) 


In  a  similar  way,  we  obtain  the  analogous  formulae: 

tan|B  =  -^-r   and   tan  |  C  =  —2-  . 
s  —  b  s  —  c 

These  formulae  will  be  referred  to  as  the  half-angle  formulae 
for  the  triangle. 

90.  Three  sides.  The  results  of  the  preceding  section  make  it 
possible  to  remove  the  one  remaining  point  contained  in  the 
criticism  of  83,  viz.,  to  devise  a  treatment  for  case  (d)  adapted  to 
calculation  by  logarithms. 

Example. 


Given,     a 

=  14.931,  b 

=  16.902,  < 

:  =  24.315. 

Required. 
Solution. 

A,  B,  C. 

a  = 
b  = 

14.931 
16.902 

c  = 

2  s  = 

24.315 

A 

56.148    and  therefore    s  = 

28.074; 

s  —  a  = 

13.143 

and 

log  0  —  a)  = 

1.11870 

s  -b  = 

11.172 

and 

log  (.s  —  b)  = 

1.04813 

s  —  c  = 

3.759 

A* 
28.074 

and 

log  (s   —  C)    — 
log  s  = 

.57507 
A 
2.74190 
1.44830 

1.29360 

*  It  is  well  to  use  this  chock  upon  the  calculation  of  the  quantities  s, 
a,  s  —  b,  and  s  —  c  before  proceeding  with  the  rest  of  the  work. 


82  INSCRIBED  AND   CIRCUMSCRIBED  CIRCLES 

Since  p  =  y/^ZSZS5Z5 1  we  haye  lQg  v  =      M68Q 

Subtracting  log  (s  —  a),  log  (s  —  b)  and  log  (s  —  c)  in  turn  from 
log  p,  we  find  by  means  of  the  half -angle  formulae: 

log  tan  A/2  =  9.52810  -  10,      log  tan  B/2  =  9.59867  -  10, 

log  tan  C/2  =  .07173; 
.-.     A/2  =  18°  38'  33",  B/2  =  21°  38'  52",     C/2  =  49°  42'  37", 

and      A  =  37°  17'  6",       £  =  43°  17'  44",       C  =  99°  25'  14". 
Check.    A  +  B  +  C  =  180°  0'  4". 

91.  Exercises. 

1.  a  =  .96834,  6  =  .94572,  c  =  .95902.     Determine  A,  B,  C. 

2.  a  =  453.67,  b  =  112.34,  c  =  369.85.     Determine  A,  B,  C. 

3.  x  =  1.004,  ?/  =  1.705,  z  =  1.526.     Determine  X,  Y,  Z. 

4.  a  =  4500,  b  =  5400,  c  =  6300.     Determine  A,  B,  C. 

5.  A  triangular  piece  of  land  is  to  be  staked  off,  so  that  its  sides  measure, 
respectively,  73.84  rods,  68.701  rods,  and  32.503  rods.  How  may  this  be 
done? 

6.  Two  vessels  leave  the  same  harbor  at  the  same  moment,  both  going 
at  the  rate  of  12  miles  per  hour.  After  2|  hours,  the  vessels  are  24  miles 
apart.  If  one  of  the  vessels  was  sailing  in  a  due  easterly  course,  in  what 
direction  was  the  other  vessel  going? 

92.  Inscribed  and  circumscribed  circles.  Area.  The  results 
of  the  preceding  articles  enable  us  to  derive  other  important 
relations,  viz.: 

(1)  A  formula  for  the  radius  of  the  inscribed  circle  (see  Fig.  45). 


The  points  of  tangency  of  the  inscribed  circle  of  radius  r  divide 
the  total  perimeter  of  the  triangle  into  six  parts,  which  are  equal 
in  pairs.     Hence  the  sum  of  the  segments  of  any  set  of  three,  of 


INSCRIBED  AND   CIRCUMSCRIBED  CIRCLES  83 

which  no  two  are  equal,  must  be  equal  to  s,  one  half  of  the  peri- 
meter; e.g., 

AF  +  BD  +  DC  =  s    or    AF  +  a  =  s; 

hence  AF  =  EA  =  s  —  a. 

In  similar  manner  we  show  that  FB  =  BD  =  s  —  b  and  DC 
=  CE  =  s  -  c. 
Moreover  A  AOF  is  a  right-angled  triangle  (Why?)  and  LOAF 

=  A/2.     (Why?) 

Therefore  tan  — 


2       AF      s  -  a 

On  the  other  hand,  we  have  from  the  half-angle  formulae  for 

the  angles  of  a  triangle :  tan  —  =  — - —  .     From  this  we  conclude 

2       s  —  a 

that  r  =  p,  i.e.: 

Theorem  VI.     The  radius  of  the  Inscribed  circle  of  a  triangle  ABC 
Is  given  by  the  formula: 


-V 


(s  -  a)  (s  -  b)  (s  -  c) 


where  s  designates  one  half  of  the  perimeter  of  the  triangle. 

(2)  A  second  formula  for  the  area  of  a  triangle. 
We  see  from  Fig.  45,  that 

A  ABC  =  A  AOB  +  A  BOC  +  A  COA 

/o    i           /o  ,        ».  /o            a  +  b  +  c 
=  r  •  c/2  -f-  ?'  •  a/2  +  r  •  0/2  =  r  • =  rs. 

Using  the  formula  for  the  radius  of  the  inscribed  circle  found 
above  we  get: 

Theorem  VII.     The  area  of  a  triangle  ABC  Is  given  by  the  formula: 


A  =  Vs  (s  -  a)  (s  -  b)  (s  -  c). 

This  formula,  known  as  Hero's  formula,  expresses  the  area  of 
the  triangle  in  terms  of  the  sides  only;  it  is  well  known  from  the 
study  of  plane  geometry.     Together  with  the  formula  derived 


84  SUMMARY  OF  RESULTS  OF  CHAPTER  VII 

in  76,  it  suffices  to  determine  the  area  of  the  triangle  in  all  cases 
which  we  have  considered. 

(3)  A  formula  for  the  radius  R  of  the  circumscribed  circle. 

We  draw  (see  Fig.  46)  the  diameter  AOD  and  connect  C  with  D. 


Fig.  46 


Then    Z  ACD  is  a  right  angle  (Why?)   and    Z  ADC  =  Z  ABC 

AC        b 
=  B  (Why?).     Hence  sin  B  =  sin  ADC  =    .-r.  =  ?m  ■     From  this 
J  '  AD      2  li 

it  follows    hat  R  =  ^—. — ^.     By  means  of  the  law  of  sines,  we 
2  sin  B 

derive  from  this    wo  other  expressions  for  R,  so  that  we  can  now 

state  the  following  theorem: 

Theorem  VIII.     The  radius  of  the  circumscribed  circle  of  a  triangle 
ABC  is  given  by  the  formulae: 

R  = 


2  sin  A      1  sin  li      2  sin  C 

93.   Summary  of  the  results  of  Chapter  VII. 

i.   Law  of  sines:  .  =  - — 75  =    .    ^  see  76. 

sm  A      sin  B      sin  C 

2.   Law  of  cosines:  a2  =  b2  +  c2  —  2  6c  cos  A,  see  81. 

52  =  c2  +  ft2  _  2  m  cos  £. 
c2  =  a2  +  52  _  2  a6  cos  C. 


SUMMARY   OF   RESULTS  OF   CHAPTER   VII  85 

3.  Law  of  tangents:  tan  h  (A  —  B)  =  — —7-  cot  h  C,  see  84. 

a  +  b 

tan  \  (B  -  C)  =  ^~  cot  \  A. 

tan|(C  -  A)  =  °^cot^B. 
c  +  a 

4.  Half -angle  formulae  for  the  triangle: 

see  89. 


tan  \  A 

V 
s—a 

tan|£ 

V 
s-b 

tan  \  C 

=     p 
s  —  c 

where  p 


{s  —  a)  (s  —  b)  (s  —  c) 


5.    Mollweide's  equations: 

a  +  b      cos  h  (A  -  B)        a-b      sin  \  (A  -  B) 


c 

siniC 

b  +  c 

cos  h{B-C) 

a 

sin  ^  A 

c  +  a 

cosHC-^) 

c 

cos  §  C 

b  -c 

sin  J  (5  -  C) 

a 

cos  h  A 

c  —  a 

sin  |  (C  -  A) 

>  see  85. 


b  sin  7*  #  b  cos  1 5 

6.    The  area  of  triangle: 

a.  A  =  \  ab  sin  C  =  |  6c  sin  yl  =  |  ca  sin  5,  see  76. 


b.    A  =  Vs  (s  -  a)  (s  -  6)  (s  —  c),  see  92. 

7.    The  radius  of  the  inscribed  circle: 


,.  =  J(s-a)(8-b)(8-c) 
V  s 

8.    The  radius  of  the  circumscribed  circle 


see  92. 


lii    ■       n     .         .  r>      '        n  n  si  '  See  i/Z. 

2  sin  yl       2  sin  #      2  sin  C 


86         MISCELLANEOUS   EXERCISES  AND   APPLICATIONS 
9.   Methods  for  solving  triangles: 


Solved  by  use  of: 

Case 

Checked  by  use  of: 

Without  logs 

With  logs 

Two  angles  and  one 

Formulae  1,  6a 

Formulae  1,  6a 

Formulae  5 

side. 

Two   sides    and   an 

Formulae  1,  6a 

Formulae  1,  6a 

Formulae  5 

angle  opposite  one 

of  them. 

Two  sides  and  the 

Formulae  2,  6a 

Formulae  3,  6a 

Formulae  5 

included  angle. 

Three  sides. 

Formulae  2,  6b 

Formulae  4,  6b 

LA  +  LB  +  LC 
=  180° 

94.   Exercises. 

1.  Prove  the  law  of  sines  by  means  of  tne  method  indicated  in  92  (3). 

2.  Show,  in  Figure  45,  that  BD  +  CE  +  EA  =  s. 

3.  Prove,  in  Figure  45,  that  CE  =  DC  =  s  -  c. 

abc 


4.   Use  92,  (3)  to  show  that  the  area  of  a  triangle  is  equal  to 


4/r 


Determine  the  area,  and  the  radius  of  the  circumscribed  circle  for  each  of 
the  following  triangles: 

5.  a  =  .473,  6  =  .586,  C  =  23°  47'  12". 

6.  x  =  3.045,  Y  =  47°  28',  Z  =  65°  34'. 

7.  a  =  41.35,  b  =  36.78,  A  =  35°  27'. 

8.  c  =  632,  a  =  571,  B  =  30°. 

Determine  the  area,  and  the  radius  of  the  inscribed  circle  for  each  of  the 
following  triangles: 

9.  a  =  40.37,  b  =  31.56,  c  =  27.08. 

10.  a  =  .971,  b  =  .506,  c  =  .683. 

11.  a  =  437,  c  =  856,  C  =  38°  41'. 

12.  a  =  .0456,  b  =  .0731,  C  =  74°  26'. 


95.   Miscellaneous  exercises  on  Chapter  VII  and  applications. 

1.  The  area  of  a  triangular  piece  of  land  is  43  acres.  One  side  measures 
440  yards,  and  the  angle  at  one  extremity  of  this  side  is  43°.  What  must  the 
remaining  sides  and  angles  be?     (1  acre  =  4840  sq.  yds.) 


MISCELLANEOUS  EXERCISES  AND  APPLICATIONS        87 

2.  A  lighthouse  is  observed  N  15°  W  from  a  vessel  which  is  sailing  15 
miles  an  hour  in  a  due  northerly  course.  Half  an  hour  later  the  bearing  of 
the  same  lighthouse  is  N  37°  W.  How  far  is  the  lighthouse  from  the  second 
position  of  the  vessel  and  how  long  will  it  be  before  the  lighthouse  is  sighted 
due  West? 

3.  From  the  top  of  a  mountain  the  angles  of  depression  of  two  consecu- 
tive milestones  in  the  same  vertical  plane  with  the  top  of  the  mountain  are 
10°  and  15°.     How  high  is  the  mountain  ? 

4.  A  forester  observes  that  the  angle  of  elevation  of  an  observation  tower 
from  a  point  P  is  5°;  after  walking  towards  the  tower  along  a  horizontal  road 
for  a  distance  of  500  feet,  he  observes  that  the  angle  of  elevation  has  changed 
to  35°.     How  much  farther  will  he  have  to  go  to  reach  the  foot  of  the  tower? 

Calculate  the  unknown  parts  and  the  area  of  each  of  the  triangles  indicated 
in  Exs.  5-7: 

5.  a  =  47.032,  b  =  35.614,  A  =  27°  45'  16". 

6.  a  =  3,  b  =  5,  c  =  7. 

7.  a  =  23.  c  =  17,  C  =  42°  23'. 

Determine  the  area  and  the  radii  of  the  inscribed  and  circumscribed 
circles  in  each  of  the  triangles  indicated  in  Exs.  8-10: 

8.  a  =  256,  C  =  17°  13',  B  =  45°  16'. 

9.  b  =  2.25,  c  =  1.75,  A  =  54°. 

10.  a  =  15,  b  =  17,  c  =  25. 

11.  The  ratio  of  the  lengths  of  two  sides  of  a  triangle  is  5  :  8,  the  included 
angle  is  35°.     Determine  the  other  angles  of  the  triangle. 

12.  A  wireless  tower  is  built  on  the  edge  of  a  cliff.  From  a  boat  at  sea, 
the  angle  of  elevation  of  the  top  of  the  tower  is  30°.  After  rowing  towards 
the  shore  for  a  distance  of  400  feet  it  is  found  that  the  angles  of  elevation 
of  the  bottom  and  top  of  the  tower  are  45°  and  57°  respectively.  How  high 
is  the  cliff  and  how  high  is  the  tower? 


200  ft 


13.  To  determine  the  distance  between  two  points  A  and  B,  situated  on 
the  surface  of  a  lake,  two  points  C  and  I)  an;  selected  in  such  a  manner  that 
both  .4  and  B  are  visible  from  C  and  1)  (see  Fig.  47).     It  is  found  that  VI) 


88         MISCELLANEOUS  EXERCISES  AND  APPLICATIONS 

=  200    feet    and    that     /ACB  =  33°,     / BCD  =  54°,     ^ADB  =  40°     and 

snr^A        «-o      ™  ,,    .     An       200  sin  65°  .   ..     .    __       200  sin  75° 

/CD  A  =  6o  .     Show  that  AC  =  : — ^^ —  ,  and  that  BC  =  — : — ^— 5 — . 

sin  28  sin  21 

Then  calculate  the  distance  AB. 

14.  A  vessel  is  sailing  due  east  at  the  rate  of  20  miles  per  hour.  At  10 
a.m.  a  lighthouse  L  is  bearing  N  10°  W,  while  a  second  lighthouse  M  is  bear- 
ing N  40°  E;  at  2  p.m.  the  bearings  of  L  and  M  are  N  50°  W  and  N  5°  E 
respectively.     How  far  are  L  and  M  apart? 

15.  Determine  also  from  the  observations  recorded  in  Problem  14  the 
direction  of  the  line  from  L  to  M. 

16.  To  measure  the  height  of  a  mountain  AF,  above  a  horizontal  plane  P, 
we  measure  the  distance  between  two  stations  B  and  C  (see  Fig.  48),  so  selected 


Fig.  48 

that  at  least  one  of  them,  say  B,  lies  in  the  plane  P,  that  each  is  visible  from 
the  other  and  that  A  is  visible  from  both.  The  angles  ACB,  ABC  and  the 
angle  of  elevation  a  of  .4  as  seen  from  B  are  measured.  It  is  found  that 
BC  =  500  feet,  /.ACB  =  85°  25',  I  ABC  =  84°  33' and  a  =  40°  17'.  Deter- 
mine the  height  of  the  mountain  above  the  plane  P. 

17.  The  angle  of  elevation  of  a  church  steeple  T  from  a  point  R  is  17°  25'. 
At  a  point  S,  250  feet  from  R,  the  line  TR  subtends  an  angle  of  73°  47',  while 
from  R  the  line  TS  subtends  an  angle  of  653  8'.  Determine  the  height  of  T 
above  the  horizontal  plane  through  R. 

18.  A  flag  staff  on  top  of  a  monument  subtends  an  angle  of  3°  at  a  point 
P,  400  feet  above  the  ground  and  at  a  horizontal  distance  of  300  feet  from 
the  foot  of  the  monument.  From  the  same  point  P  the  monument  itself  sub- 
tends an  angle  of  43°.  Determine  the  height  of  the  flag  staff  and  the  height 
of  the  monument. 

19.  One  side  of  a  triangle  is  75  feet  and  the  angle  opposite  this  side  is  34°. 
The  sum  of  the  other  sides  is  125  feet.  Determine  all  the  sides  and  angles  of 
this  triangle. 

20.  A  building  20  feet  high  is  surmounted  by  a  steeple  30  feet  high.  How 
far  from  the  foot  of  the  building  must  an  observer  stand  in  order  that  the 
building  and  steeple  may  subtend  the  same  angle  at  his  eye,  which  is  5  feet 
above  the  grtuind? 

21.  A  pole  10  feet  tall  is  divided  into  two  parts,  a  lower  part  of  6  feet  and 
an  upper  part  of  4  feet.      Frtnn  a  point  P.  7  feet  above  the  bottom  of  the  pole, 


MISCELLANEOUS   EXERCISES   AND   APPLICATIONS 


89 


these  two  parts  subtend  the  same  angle,  the  pole  being  held  vertically.     How 
far  is  the  pole  from  P? 

22.  The  angle  of  elevation  of  a  church  steeple  from  a  point  A,  due  south 
of  it,  is  27°;  and  from  a  point  B,  due  west  of  the  steeple,  and  in  the  same  hori- 
zontal plane  as  A,  the  angle  of  elevation  is  35°.  Moreover  the  distance  AB 
is  150  yards.     Determine  the  height  of  the  steeple  above  the  plane  AB. 

23.  From  a  point  A  the  angle  of  elevation  of  the  top  T  of  a  flagpole  which 
stands  on  top  of  a  building  is  37°  47'.  From  a  point  B,  100  feet  nearer  to  the 
building  and  lying  in  the  vertical  plane  through  T  and  A,  the  angle  of  eleva- 
tion of  T  is  47°  32'.     How  high  is  T  above  the  ground? 

24.  The  angle  of  elevation  of  the  bottom  of  the  flagpole  described  in 
Problem  23,  as  seen  from  B,  is  45°  25'.     Determine  the  height  of  the  pole. 

25.  To  determine  the  height  XY  of  a  wireless-tower  X  above  a  horizontal 
plane  P  two  points  A  and  B  are  selected  in  this  plane  P  (see  Fig.  49).     The 


Fig.  49 


angles  XAY,  XBY,  YAB  and  ABY  and  the  distance  AB  are  measured.  It 
is  found  that  Z  XAY  =  67°,  Z  XBY  =  58°  3'.  1,  Z  YAB  =  27°,  Z  YBA  =  1SC 
and  AB  =  81.6  feet.     Determine  the  height  XY  and  check  the  calculations. 


CHAPTER  VIII 

INVERSE   TRIGONOMETRIC   FUNCTIONS.     TRIGONO- 
METRIC  EQUATIONS 

96.  Inverse  functions.  The  concept  function  which  we  have 
used  in  preceding  chapters  will  now  be  studied  in  somewhat 
greater  detail.  The  concept  may  be  denned  for  our  purpose  in 
the  following  way: 

Definition  I.  If  a  relation  between  two  variables,  x  and  y,  is  given 
in  such  a  way  that  to  every  value  of  either  there  correspond  one  or 
more  values  of  the  other,  then  .t  is  a  function  of  v,  and  y  is  a  function 

of  X. 

If  these  two  functional  dependences  of  y  upon  x  and  of  x  upon 
y  are  written  down  explicitly,  we  obtain  two  functions,  one  in 
terms  of  x,  yielding  y,  the  other  in  terms  of  y,  yielding  x.  In 
the  first  of  these  functions  x  is  the  independent  variable  and  y  the 
dependent  variable;  in  the  second  y  is  the  independent  variable 
and  x  the  dependent  variable.  Two  such  functions  which  result 
from  the  same  relation  between  the  two  variables  are  called 
inverse  with  respect  to  each  other.  This  may  be  expressed  as 
follows : 

Definition  II.  If  a  relation  between  two  variables,  x  and  y,  be 
solved  in  turn  for  r  in  terms  of  y  and  for  y  in  terms  of  .r,  we  obtain 
two  functions  which  form  a  pair  of  inverse  functions. 

The  important  connections  between  them  are  recognized  more 
readily  if  one  letter  be  used  for  the  independent  variable  in  both 
functions  and  another  letter  for  the  dependent  variable  in  both 
functions. 

Example  1.  The  relation  between  the  variables  .r  and  y  de- 
termined by  the  equation 

3x  +  2?y-7  =  0     (1) 
gives  rise  to  the  functions 

7  —  3x     /n  .  7  —  2  y     ,_. 

y  =  — - —     (2)     and     x  =  — ^— "-     (3). 

90 


INVERSE  FUNCTIONS 


91 


The  graphical  representation  of  the  relation  between  x  and  y 
can  readily  be  obtained  from  equations  (2)  or  (3);  in  either  case 
we  will  obtain  the  straight  line  of  Figure  50. 


)\ 

J 

3 

\ 

2 

-      \ 

1 

O 

1              2   2H        3 

Fig.  50 


Denoting  the  independent  variable  by  t  and  using  u  to  denote 
the  dependent  variable  in  both  cases,  equations  (2)  and  (3)  will 
become 


u  — 


Zt  ,  7-2* 

and     u  —  — ~ 


3? 

<u 

21s 

\  u  » 

7-3  t 

2 

1 

1 

<T 

>  T 

O 

1 

2\      3    3fS.    4 

Fig.  51 


Here  we  have  a  pair  of  inverse  functions,  whose  graphs  are  repre- 
sented in  Figure  51. 


92 


GRAPHS  OF  INVERSE  FUNCTIONS 


Example  2.     The  relation  y2  —  x  =  0  gives  rise  to  the  pair  of 
inverse  functions 

u  =  t2    and     u  =  vt, 
whose  graphs  are  given  in  Figure  52. 

u 

\ 


Fig.  52 

In  the  first  function  of  this  pair,  one  'positive  value  of  u  corre- 
sponds to  every  value  of  t;  but  in  the  other  function  two  values  of  u 
correspond  to  every  positive  value  of  t,  and  no  value  of  u  to  negative 
values  of  t.  The  function  u  —  t2  is  a  single-valued  function  of  t, 
defined  for  all  values  of  t;  the  inverse  function  u  =  x^t  is  a  two- 
valued  function  of  t  defined  for  positive  values  of  t  only.  It  may 
happen  that  the  inverse  function  of  a  single-valued  function  is 
3-valued,  or,  in  general,  multiple-valued. 

97.  Graphs  of  inverse  functions.  Let  the  functions  u  =  f  (t) 
and  u  =  g  (t)  be  a  pair  of  inverse  functions  whose  graphs  are  the 
curves  drawn  in  Figure  53.  If  the  point  P  (a,  b)  belongs  to  the 
graph  of  u  =  f  (t),  then  the  point  Q  (b,  a)  must  be  on  the  graph  of 
u  =  g  {t),  because  the  two  functions  may  be  thought  of  as  having 
been  obtained  from  one  and  the  same  equation  between  x  and  y, 
x  and  y  being  replaced  by  t  and  u  in  one  case,  and  by  u  and  t  in 
the  other.  Now  it  is  readily  seen  that  two  such  points  P  (a,  b) 
and  Q  (6,  a)  are  symmetrically  situated  with  respect  to  the  45° 
line  MN  (see  Fig.  54),  i.e.,  that  MN  is  the  perpendicular  bisector 
of  the  line  PQ.     For  As  ROP  and  QOS  are  congruent,     (Why?) 


GRAPHS  OF  INVERSE  FUNCTIONS 


93 


Hence  PO  =  QO  and  Z  ROP  =  Z  SOQ.  Consequently  Z  POT 
=  Z  QOT  (Why?)  and  A  POT  and  A  QOT  are  congruent. 
(Why?)    Therefore  PT  =  QT  and  Z  PTO  =  Z  QTO  =  90°,  which 


*-r 


Fig.  53 


I 
s 
a 

/!  X 

>  r 

/ 

O             a        ft 

Fig.  54 


was  to  be  proved.  Consequently  from  points  A,  B  on  the  graph 
of  any  function,  points  A',  B' ,  ...  on  the  graph  of  the  inverse 
function  may  be  obtained  by  determining  A',  B',  ...  as  points 
which  are  symmetrically  situated  with  .1,  B,  .  .  .  with  respect 
to  the  45°  line  MN,  i.e.,  by  reflecting  the  points  A,  B,  .  .   .  in 


94 


THE  INVERSE  SINE  FUNCTION 


MN  as  a  mirror*.     These  results  may  be  summarized  in  the  fol- 
lowing theorem: 

Theorem  I.  If  u  =/  (t)  and  u  =  g  (t)  are  a  pair  of  Inverse  func- 
tions the  graph  of  either  function  may  be  obtained  by  reflecting  the 
graph  of  the  other  function  in  the  45°  line  as  a  mirror. 

98.   Exercises. 

Determine  the  inverse  function  associated  with  each  of  the  following  func« 
tions: 


1.   u  =  4  t  -  3. 


<2/3. 


3.   u  =  logio  t. 


4.    u  = 


Construct  the  graphs  of  the  following  pairs  of  inverse  functions: 
5.    u  =  Vt  -4,     u  =  t1  +  4.  6.   u  =  t/3,     u  =  3  t. 

7  At  -7 


7.   u  =  V9  -  t2,     u  =  Vo  -  t2. 


8.   u  =  V 


bt  -4' 


5<  -3 


99.  The  inverse  sine  function.  The  equation  y  =  sin  x  estab- 
lishes a  relation  between  the  variables  x  and  y,  represented 
graphically  by  the  sine  curve  (see  dashed  curve  in  Fig.  55).     For 


Fig.  55 


every  value  of  x  there  is  one  corresponding  value  of  y.  If  the 
('filiation  be  solved  for  x  in  terms  of  y,  and  x  and  y  be  replaced  by 
u  and  /  respectively,  we  obtain  the  inverse  sine  function;  it  is 
expressed  in  the  form  u  =  arc  sin  t,  which  is  read  "u  is  the  angle 
*  See  footnote  on  page  45. 


THE  OTHER   INVERSE  TRIGONOMETRIC   FUNCTIONS     95 

whose  sine  is  t."  The  graph  of  the  inverse  sine  function  of  t,  i.e., 
of  arc  sin  t,  may  be  obtained,  in  virtue  of  Theorem  I,  by  reflecting 
the  sine  curve  in  the  45°  line  as  a  mirror.  In  this  way  we  obtain 
the  full-drawn  curve  of  Figure  55. 

The  inverse  sine  function  of  t  cannot  be  expressed  in  a  simple 
manner  in  terms  of  algebraic  or  trigonometric  functions  of  t.  It 
must  be  regarded  as  an  entirely  new  function,  defined  as  the  in- 
verse of  the  sine  function.  From  this  definition  its  properties 
will  be  derived  by  the  aid  of  its  graph. 

We  observe  that  the  function  u  =  arc  sin  t  gives  an  indefinitely 
large  number  of  values  of  u  for  every  value  of  t,  which  lies  between 
—  1  and  1,  and  that  it  gives  no  value  of  u  at  all  for  values  of  t 
which  lie  outside  the  range  (  —  1,  1).  The  inverse  sine  function 
is  an  infinitely  multiple-valued  function,  defined  only  for  values  of 
t  between  —1  and  1. 

So,  e.g.,  arc  sin  \  =  30°,  150°,  510°,  -210°,  .  .  .  just  as  sin  30 
=  sin  150°  =  sin  510°  -  sin  (-210°)  =   •  •  •    =  \. 

Definition  III.  The  symbol  Arc  sin  t,  called  the  principal  value 
of  arc  sin  t,  is  used  to  designate  the  numerically  smallest  angle 
whose  sine  is  equal  to  t. 

For  example  Arc  sin  \  =  30°  =  ir/6. 

It  is  clear  that  Arc  sin  t  is  a  single-valued  function  of  t,  defined 
for  values  of  t  between  —1  and  +1.  Its  graph  is  given  by  the 
heavily  drawn  portion  of  the  graph  of  arc  sin  t  in  Figure  55. 

When  t  is  between  0  and  1,  Arc  sin  t  will   be  between  0  and  -; 

when  t  is  between  —  1  and  0,  Arc  sin  t  will  be  between  —  -  and  0. 

100.  The  other  inverse  trigonometric  functions.  In  a  similar 
manner  the  equations  u  =  arc  cos  t,  u  =  arc  tan  t,  u  =  arc  cot  t, 
u  =  arc  sec  t  and  u  =  arc  cosec  t  represent  the  inverse  functions 
associated  with  the  remaining  trigonometric  functions  cos  t,  tan  t, 
cot  t,  sec  t,  and  cosec  /  respectively.  These  equations  are  read 
"u  is  the  angle  whose  cosine  is  equal  to  t,"  etc.  The  graphs  of 
the  inverse  trigonometric  functions  are  obtained  from  the  graphs 
of  the  associated  trigonometric  functions  by  the  use  of  Theorem  I, 
i.e.,  by  reflecting  the  latter  in  the  45°  line  as  a  mirror.  In 
Figures  56  and  57,  the  graphs  of  the  functions  arc  cos  /  and  arc 
tan  /  arc  indicated  by  the  full-drawn  curves.  We  observe  thai 
both  of   these  functions  are  infinitely  multiple-valued   for  those 


96     THE  OTHER   INVERSE  TRIGONOMETRIC   FUNCTIONS 


/ 

u 

M=arc  cos 

\ 

s2 

s 

/ 

/v 

\ 
\ 

-7T                    AVi-1 
\                               '     2 

A 

-i 

-IT 

2 

V 

-7T 

Fig.  56 


u  ■  arc  tan  t 


Fig.  57 


EXERCISES  97 

values  of  the  independent  variable  for  which  they  exist  at  all;  i.e., 
for  every  value  of  t  between  —1  and  1  there  is  an  infinitely  large 
number  of  values  of  arc  cos  t;  while  for  every  value  of  t  there  is 
an  infinitely  large  number  of  values  of  arc  tan  t. 

If  cot  u  =  t,  then  tan  u  =  1/7;  hence  arc  cot  t  =  arc  tan  1/t, 
i.e.,  an  angle  whose  cotangent  is  equal  to  t  is  equal  to  an  angle 
whose  tangent  is  equal  to  1/t.  For  an  analogous  reason  we  have 
arc  sec  t  —  arc  cos  1/t  and  arc  cosec  t  =  arc  sin  1/t.  Since  the 
functions  arc  cot  t,  arc  sec  t,  and  arc  cosec  t  are  so  readily  expres- 
sible by  means  of  the  functions  arc  tan  t,  arc  cos  t,  and  arc  sin  t 
respect ively,  we  shall  limit  our  study  to  the  latter  three  functions. 

As  in  the  case  of  arc  sin  t,  we  define  single-valued  functions 
corresponding  to  the  multiple-valued  functions  arc  cos  t  and  arc 
tan  t. 

Definition  IV.  The  symbol  Arc  cos  t,  called  the  principal  value 
of  arc  cos  t,  designates  the  least  positive  angle  whose  cosine  is  equal 
to  t. 

Definition  V.  The  symbol  Arc  tan  t,  called  the  principal  value 
of  arc  tan  t,  designates  the  numerically  smallest  angle  whose  tangent 
is  equal  to  t. 

The  graphs  of  the  single-valued  functions  Arc  cos  t  and  Arc 
tan  t  are  given  by  the  heavily  drawn  portions  of  Figures  56  and 
57  respectively. 

When  t  lies  between  0  and  1,  Arc  cos  t  is  between  0  and  7r/2; 
when  t  lies  between  —  1  and  0,  Arc  eos  /  is  between  t/2  and  t; 
when  t  is  positive,  Arc  tan  t  is  between  0  and  7r/2; 
when  /  is  negative,  Arc  tan  t  is  between  —  tt/2  and  0. 

101.   Exercises. 

1.  Determine  arc  cos  l\  Arc  tan  1;  Arc  cot  ^3:  Arc  cos  1;  arc  sin  0;  Arc 
cos  1    V2. 

2.  Evaluate  sin  (arc  sin  \);  cos  (arc  co.s  §);   tan  (arc  tan  V7). 

3.  Determine:  cos  ('arc  tan  1/V3);  tan  (arc  sin  v£);  sec  (arc  cos  J); 
sin  (arc  cos  .4321);    cos  (Arc  sin  0);    cot  (Arc  tan  Vs). 

4.  Evaluate :_ cos  (Arc  cos  ]  +  Arc  sin  1);  tan  (Arc  sin  1/V2  —  Arc  tan  1); 
sin  (Arc  tan  \73  +  Arc  cos  1  /\A2\ 

5.  Construct  the  graphs  of  the  functions  u  =  arc  sin  2  /;  u  =  arc  sin  /  .'!; 
u  =  arc  cos  t/'i. 


98       MULTIPLE-   AND  SINGLE-VALUED   INVERSE   FUNCTIONS 

102.  Relations  between  multiple-valued  and  single-valued  in- 
verse functions.  There  is  a  simple  algebraic  relation  between 
the  multiple-valued  function  arc  sin  t  and  the  single-valued  func- 
tion Arc  sin  t.     If  u  =  Arc  sin  t,  we  know  that 

t  =  sin  u  =  sin(w  •  180°  +  u),  when  n  is  an  even  integer,  and  that 
t  =  s'mu  =  sin(n  •  180°  —  u),  when  nis  an  odd  integer  (see  Chap- 
ters V  and  VI). 

Hence      arc  sin  t  =  n  •  180°  +  Arc  sin  /,  when  n  is  even, 
and  arc  sin  t  =  n  •  180°  —  Arc  sin  t,  when  n  is  odd. 

These  two  equalities,  which  may  be  derived  directly  from  the 
graph  of  arc  sin  t  (see  Fig.  55) ,  are  combined  in  the  single  formula : 

arc  sin  t  =  n  •  180°  +  (-1)"  Arc  sin  t. 

In  entirely  analogous  manner  we  obtain  the  further  formulae: 

arc  cos  t  =  n  •  360°  ±  Arc  cos  t, 
arc  tan  t  =  n  •  180°  +  Arc  tan  t. 

103.  Trigonometric  equations.  Let  us  consider  the  following 
problem:  To  construct  a  rectangle  whose  perimeter  is  10"  and 
whose  diagonal  is  4". 


Fig.  58 

If  A  BCD  be  the  required  rectangle  and  6  the  angle  between  the 
diagonal  AC  and  the  side  AB  (see  Fig.  58).  we  have: 

AB  =  AC  cos  0  =  4  cos  0 

and 

BC  =  AC  sin  6  =  4  sin  0. 

The  conditions  of  the  problem  are  therefore  expressed  by  the 
equation 

8  cos  9  +  8  sin  6  =  If). 

This  equation  involves  trigonometric  functions  of  the  unknown 
6;  it  is  called  a  trigonometric  equation.     Equations  of  this  sort 


SOLUTION   OF  TRIGONOMETRIC  EQUATIONS  99 

occur  in  various  branches  of  mathematics  and  in  its  applications. 
We  consider  their  solution. 

104.  Solution  of  trigonometric  equations.  "  To  solve  a  trigo- 
nometric equation"  means  to  determine  all  possible  angles  which 
satisfy  the  equation.  It  is  important  to  notice  the  difference 
between  a  trigonometric  identity,  which  is  true  for  all  angles  for 
which  it  has  a  meaning,  and  a  trigonometric  equation,  which  is 
satisfied  by  certain  angles  only.  The  determination  of  these 
special  angles  constitutes  the  "solution  of  the  trigonometric 
equation." 

The  work  of  solving  a  trigonometric  equation  may  be  divided 
into  three  parts,  viz. : 

(1)  To  express  all  the  trigonometric  ratios  occurring  in  the 
equation  in  terms  of  a  single  ratio  of  a  single  angle, 

(2)  To  solve  the  resulting  equation  obtained  in  (1),  as  an 
algebraic  equation  for  that  ratio,  and 

(3)  To  determine  all  the  angles  corresponding  to  the  values 
for  that  ratio  found  in  (2) . 

Returning  to  the  example  of  103,  we  proceed  as  follows: 

(1)  We  express  cos  0  in  terms  of  sin  0  by  means  of  the  formula: 
cos  0  =  Vl  —  sin2  0.  Substituting  this  in  the  equation  derived 
in  103,  we  obtain  the  equation 

sin  0  +  Vl  -  sin2  0  =  f , 
or 

Vl  -  sin2  0  =  i  -  sin  0. 

(2)  We  square  both  sides  of  this  equation,  collect  terms  and 
clear  of  fractions.  In  this  way  we  obtain  the  following  quadratic 
equation  in  terms  of  sin  0: 

32  sin2  0  -  40  sin  0  +  9  =  0. 

This  equation  we  solve  by  means  of  the  quadratic  formula, 
which  gives  us: 

40  ±  VlGOO  -  1152       40  ±  V448       40  ±  21.1660 


sm  0 


64  64  64 

sin  0  =  61.1660/64     or     18.8340/64. 


(3)  From  this  we  conclude: 

0  =  arc  sin  61.1660/64     or     arc  sin  18.8340/64. 


100  FURTHER   EXAMPLES 

By  means  of  the  tables,  we  find 

Arc  sin  61.1660/64  =  72°  53'  and  Arc  sin  18.8340/64  =  17°  7'. 
.-.     6  =  n  •  180°  +  (-1)"  72°  53'  or  n  •  180°  +  (-1)"  17°  7'. 

The  nature  of  the  problem  is  such  that  only  acute  angles  9 
are  applicable,  so  that  we  need  only  consider  the  principal  values 
of  9.     Using  these,  we  find : 

AB  =  4  cos  9  =  1.177  or  3.823, 

and  BD  =  4  sin  9  =  3.823  or  1.177; 

we  verify  that        2 AB + 2 BD  =  10. 

105.   Further  examples. 

1.    To  solve  the  equation 

tan  9  •  tan  2  9  +  cot  9  +  2  =  0, 
we  use  the  formulae: 

(1)         tan  2  6  =  t  2  ^n  i  a     and     cot  9  =  . ■  l—       (see  74,  9). 

1  —  tan-  6  tan  9 

Substituting  these  expressions  in  the  original  equation,  it 
becomes : 

2  tan2  9 1_       9  = 


1  -  tan2  9       tan  9 

(2)  This  is  an  algebraic  equation  in  terms  of  tan  9;  we  clear 
the  equation  of  fractions  and  collect  terms.  This  leads  us  to  the 
following  quadratic  equation: 

tan2  9-2  tan  0-1=0. 

If  we  solve  this  equation  for  tan  9  by  means  of  the  quadratic 
formula,  we  find: 


tan  9  =  o  =  1  -  =  2"4142  or  --4142- 

(3)  It  remains  to  determine  arc  tan  2.4142  and  are  tan  —.4142. 
From  the  tables,  we  find:  Are  tan  .4142  =  22°  30',  and  therefore 
Are  tan  (-.4142)  =  -22°  30',  whence  follows  are  tan  (-.4142) 
=  n-  180°  -22°  30'. 


FURTHER  EXAMPLES  101 

Furthermore  Arc  tan  2.4142  =  67°  30',  and  therefore  arc  tan 
2.4142  =  n  •  180°  +  67°  30'. 

2.  To  solve  the  equation 

3  sin2  6  -  4  sin  0  -  4  =  0. 

In  this  problem  the  object  of  step  (1)  has  already  been  accom- 
plished, since  the  equation  in  its  original  form  is  an  algebraic 
equation  in  terms  of  sin  0. 

Solving  it  by  means  of  the  quadratic  formula,  we  find: 
sin0  =  2  or  -§  =  -.6667. 

There  are  no  angles  whose  sine  equals  2 ;  we  have  only  to  deter- 
mine therefore  arc  sin  (  —  .6667).  From  the  tables,  we  find  that 
Arc  sin  .6667  =  41°  49';  hence  Arc  sin  (-.6667)  =  -41°  49', 
from  which  we  obtain  the  final  result: 

0  =  arc  sin  (-.6667)  =  n  •  180°  +  (-1)'  •  (-41°  49'). 
For:  n  =  1,  we  find       6  =  180°  +  41°  49'  =  221°  49'; 
n  =  2,  6  =  360°  -  41°  49'  =  318°  11'  ; 

n  =  -1,  0  =  -180°  +  41°  49'  =  -138°  11'; 

n  =  -2,  0  =  -360'  -  41°  49'  =  -401°  49'; 

n  =  0,  0  =  -41°  49';  etc. 

The  method  outlined  above  for  the  solution  of  trigonometric 
equations  is  not  always  the  shortest  or  the  most  convenient  one. 
Frequently  other  special  methods  enable  us  to  obtain  a  solution 
more  quickly  or  in  better  form.  The  student  will  do  well  to  bear 
this  in  mind,  particularly  in  cases  in  which  the  algebraic  equa- 
tion, to  which  the  method  described  above  leads,  is  such  that  he 
cannot  solve  it.  One  such  special  method  is  illustrated  by  the 
following  example. 

3.  To  solve  the  equation 

cos  2  x  —  sin  2  x  =  1 . 

Since  tan  45°  =  1,  we  may  write  this  equation  in  the  form 
tan  45°  cos  2  x  —  sin  2  . r  =  1 , 

from  which  we  derive,  by  multiplying  both  sides  of  the  equation 
by  the  corresponding  sides  of  the  equality  cos  45°  =  1  /\/2, 
sin  45°  cos  2x  —  cos  45°  sin  2  x  =  1  'vA2, 

or  sin  (45°  -  2  x)  =  1/^2. 


102  FURTHER    EXAMPLES 

From  this  we  conclude  that  we  must  have 

45°  -  2  x  =  arc  sin  1/V2  =  n  •  180°  +  (-1)"  •  45°, 
i.e.,  2  x  =  45°  -  n  •  180°  -  (-1)"  •  45°, 

or  finally:  x  =  45°  ~  (~l)n-4g;  _  n  §  g()0> 

For  n  =  0,  1,  —  1,  2,  —2,  we  obtain  the  special  values 
3  =  0°,  x  =  -45°,  x  =  135°,  x  =  -180°,  a;  =  180°  respec- 
tively. 

The  same  method  may  be  applied  in  solving  any  equation  of 
the  form 

a  cos  m  6  +  6  sin  m  6  =  c, 
provided 


\c\  <  Va'+b\ 

We  divide  both  sides  of  the  equation  by  b  and  determine  the 
angle  a  by  the  formula  a  =  Arc  tan  a/6.     Then  tan  a  =  a/b,  and 

cos  a  =  ,  the  square  root  being  taken  with  the  sign  of  6; 

v^  +  62 
then  the  equation  will  take  the  following  form: 

tan  a  cos  m  8  +  sin  m  6  =  c/b. 

We  now  multiply  both  sides  of  the  equation  by  cos  a  and  make 
use  of  the  addition  formula  for  the  sine;   in  this  manner  we  find 

.       .  nS         C  COSa  C 

sin  (a  +  m  6)  =  — r—  -      ,  „        ;/ 
6  v  a '  +  6- 

Since  now  'c1  S  Va2  +  62,  the  right  hand  side  of  this  equation  is 
always  numerically  less  than  1.  It  may  therefore  be  solved  for 
a  +  m  6,  so  that  we  obtain  the  final  result: 


=  -   \  -a  +  arc  sin      ,=  = 
m  L  \'o-  +  6-J 


where  a  =  Arc  tan  a/b,  and  where  the  square  root  is  to  be  taken 
with  the  sign  of  6. 


EXERCISES  103 

106.   Exercises. 

1.  Solve  the  equation:  tan  x  +  3  cot  x  =  4. 

2.  Solve  the  equation:  sin  0  -f  cos  0  =  V2/2. 

3.  Solve  the  equation:  tan  0  —  3  cos  0  +  sec  0  =  0. 

4.  Solve  the  equation:  cos  2  a;  =  3  sin  x  +  2. 

5.  Solve  the  equation:  sin  0  =  tan  0/2. 

6.  Solve  the  equation :  cos  2  A  =  2  sin2  ^1 . 

7.  Solve  the  equation:  4  sec2  0  =  3  (tan  0  +  2). 

8.  Determine  all  the  angles  whose  tangent  is  the  reciprocal  of  their  sine. 

9.  Determine  all  the  angles  whose  secant  is  the  reciprocal  of  their  tangent. 

10.  The  graphs  of  the  functions  sin  0  and  tan  0  are  drawn  with  respect  to 
one  set  of  axes  and  units.  Determine  all  those  values  of  0  for  which  the 
ordinate  on  the  tangent  curve  is  twice  as  long  as  the  corresponding  ordinate 
on  the  sine  curve. 

11.  Construct  an  isosceles  triangle  whose  perimeter  is  20  inches  and 
whose  altitude  is  6  inches.  (Hint.  Take  the  base  angle  of  the  triangle  as 
the  unknown.) 

12.  Construct  a  right  triangle  of  which  the  perimeter  is  36  inches  and  one 
leg  is  12  inches. 

13.  Determine  a  point  P  on  the  circumference  of  a  circle  of  radius  5  inches, 
the  sum  of  whose  distances  from  the  extremities  of  a  fixed  diameter  A  B  is 
equal  to  14  inches. 

14.  Construct  a  triangle  ABC  such  that  the  perimeter  is  equal  to  30 
inches,  the  side  c  is  equal  to  10  inches  and  the  angle  B  is  equal  to  60°. 

15.  Resolve  a  force  of  100  pounds  into  two  mutually  perpendicular  com- 
ponents, whose  sum  is  equal  to  120  pounds. 

16.  Construct  a  triangle  ABC  of  which  the  perimeter  is  60  feet  the  side 
a  is  equal  to  20  feet  and  the  angle  B  is  equal  to  45°. 


ANSWERS 

Art.  4,  Page  3 
2.   (3,3  V3).    3.    (2,-2).     6.    (0,0).     7.    (3,5);  (5,0);   (1,1).    8.  5,  Vfl, 

V52,  5  y/2.     9.    ±4;  ±5;  impossible;  ±Vs. 

Art.  9,  Page  6 

2.  65°,  12°,  113°,  -64°,  -127°,  202°,  -235°,  -337°,  598°.  3.  91°,  53°, 
392°,  -15°,  495°,  -107°,  -333°,  639°. 

Art.  11,  Page  8 

1.  x/4,  x/6,  -5  x/4,  11  x/6,  3x,  -5  x/6,  x/3,  -3  tt/2,  -9x/2,  7  x/6, 
8  tt/3,  -65  tt/18.  2.  270°,  -240°,  72°,  225°,  -90°,  270°/x,  414  °/x.  6. 
300  x.     6.   10  feet.     7.    144 °/x.     8.   2  feet.     9.   6000  x.     10.   .84  feet. 

Art.  22,  Page  14 
2a.     1.       26.    -V§/2.    2c.   0.     2d.    (-3  +  V3)/4.     3a.     -7^2/12.     36. 
3,2.     3c.    (-2  +  V3)/3,     3d.   46/9. 

Art.  25,  Page  17 
3a.    -13/4.     36.   4.     3c.    -I/V3.     3d.    -6. 

Art.  28,  Page  19 

2.  sine  =  ±  4/Vl7,  cose  =±1/Vl7,  cote  =  1/4,  sec  0  =  ±Vl7,  cosec  9 

=  ±v/l7/4. 

Art.  32,  Page  22 

1.   3,  -3,  1/3,  0,  3/5,  1.     4.   2/3,  3/2,  3/2,  -2/3,  -1/4. 

Art.  34,  Page  23 
8.    .46831.     9.     .28452.     10.   2.21638.     11.    .60061.     12.    -.93404. 

Art.  39,  Page  30 

1.  3.1335.  2.  1.9851.  3.  .0052026.  4.  5712.  5.  .63014.  6.  4.2266. 
7.  583.35.  8.  8.524.  9.  39.478.  10.  .31365.  11.  2.0550.  12.  1.3425. 
13.  4.0198.  14.  7360.5.  15.  .38638.  16.  30.628.  17.  .0093403.  18. 
.0602s.-,.     19.   .31(197.     20.  .1.3321. 

105 


106  ANSWERS 

Art.  44,  Page  35 

1.  A  =  35°  20'  28",  B  =  54°  39'  32",  c  =  644.83.  2.  A  =  62°  10'  8", 
£  =  27°  49'  52",  c  =  .16193.  3.  A  =  30°,  B  =  60°,  6  =  3.717.  4.  A  = 
48°  10'  51",  B  =  41°  49'  9",  a  =  14.609.  5.  B  =  21°  35',  a  =  22,073,  c  = 
23,737.  6.  A  =  62°  47',  a  =  .12279,  c  =  .13808.  7.  B  =  5°  25',  a  = 
376.32,  6  =  35.682.  8.  B  =  45°  25',  a  =  265.33,  b  =  269.22.  9.  £  =  44° 
57',  b  -  .08497,  c  =  .12027.  10.  A  =  49°  56'  5",  B  =  40°  3'  55",  b  = 
.057956.  11.  A  =  16°  14'  26",  B  =  73°  45'  34",  c  =  14.252.  12.  A  = 
84°  58',  a  =  1.0020,  b  =  .088254.  13.  F  =  65°  34',  a  =  24.751,  h  =  20.809. 
14.  P  =  25°  37',  a  =  19.940,  h  =  8.621.  15.  P  =  41°  51'  32",  V  =  96° 
16'  56",  fe  =  17.950.  16.  b  =  .84246,  a  =  1.1038,  P  =  67°  34'.  17.  P  = 
53°  34',  V  =  72°  52',  a  =  20.368.  18.  b  =  .64065,  a  =  .32559,  F  =  159° 
22'.  19.  P  =  62°  49'  45",  V  =  54°  20'  30",  /i  =  8.8480.  20.  b  =  12.286, 
h  =  2.9235,  7  =  129°  6'.     21.    b  =  23,454,  h  =  11,727,  P  =  45°. 

Art.  47,  Page  38 

1.   4.3301  feet.     2.   2.5  feet.     3.   0  feet;   7  feet, 

Art.  49,  Page  39 

1.  4,504.8  feet;  4,247.7  feet,  2.  272.04  feet.  3.  2,012.2  feet.  4. 
2,641.4  feet.  5.  27.155  feet.  6.  52.781  miles;  72.705  miles.  7.  53.868  feet; 
23.246  feet.  8.  1.3313  miles.  9.  566.09  feet;  504.38  feet.  10.  90.21  feet; 
125.30  feet.  11.  7.980  miles.  12.  .059  miles.  13.  320.43  feet;  640.86  feet. 
14.   8f  feet;  28  feet.     15.    67.127  miles;  78.142  miles;    104.43  miles. 

Art.  67,  Page  67 
1.    (V6  +  V2)/4,  (Vg  -  V2)/4. 

Art.  69,  Page  69 
1.  2  +  V3,  2  -  V3. 

Art.  71,  Page  61 


3.    V8-2  V6-2  V2/4,  Vs  +  2  V6  +  2  V2/4,  (V2  -  1)  (Vs  -  V2). 

Art.  73,  Page  62 

1.  2  sin  60°  cos  5°.  2.  -2  sin  45°  sin  30°.  3.  2  sin  4°  30'  cos  22°  30'. 
4.  2  cos  64°  30' cos  22°  30'.  5.  2  cos  12°  sin  30°.  6.  -2  sin  45°  cos  12°. 
7.   .22490.     8.   2.8834.     9.   -  .24629.     10.   .99726. 

Art.  76,  Page  63 

la.  (Vo-v/2)/4,-(V6+V2)/4,V3-2.  lb.  V2-V3/ 2,-^2+^3/2, 
V3-2.     4.     .14762. 


ANSWERS  107 

Art.  78,  Page  67 

1.  a  =  .29015,  c  =  .23995,  C  =  31°  42'.  2.  a  =  173.95,  b  =  237.62,  fi  = 
105°.  3.  a  =  12.736,  6  =  6.6607,  A  =  62°  55'  25".  4.  b  =  170.27, 
c  =  145.80,  fi  =  84°  20'  50".     6.    559.40  feet.     7.    137.18  feet. 


Art.  80,  Page  73 

1.  c  =  72.612,  B  =  51°  40'  30",  C  =  92°  44'  30";  c'  =  20.149,  5'  =  128° 
19'  30",  C  =  16°  5'  30".  2.  a  =  1187.5,  .1  =  140°  23'  56",  C  =  15°  49' 
4".  3.  No  solution.  4.  6  =  5.1662,  A  =  16°  7'  27",  5  =  26°  34'  33". 
5.  c  =  79.735,  B  =  26°  15'  36",  C  =  134°  19'  24";  c'  =  13.280,  #'  =  153° 
44'  24",  C"  =  6°  50'  30".  6.  a  =  .082050,  A  =  122°  58',  B  =  28°  31'- 
7.   100.91  miles.     8.   203.87  feet  or  195.72  feet.     9.   61.671  miles. 


Art.  82,  Page  76 

1.  c  =  137.48,  A  =  49°  6'  22",  B  =  70°  53'  36".  2.  P  =  53°  7'  50", 
Q  =  59°  29'  25",  R  =  67°  22'  50".  3.  A  =  26°  44',  B  =  23°  23'  12",  C  = 
129°  52'  48".  4.  t  =  .56081,  R  =  34°  34'  6",  S  =  100°  26'.  5.  399  feet. 
6.   84°  15'  37". 


Art.  88,  Page  80 

1.  a  =  97.988,  B  =  52°  31'  18",  C  =  14°  44'  42".  2.  b  =  5247.1,  A  = 
36°  54'  20",  C  =  54°  38'  40".  3.  r  =  .04343,  P  =  47°  16'  5",  <2  =  85°  7' 
55".  4.  y  =  1.2136,  A'  =  35°  28'  24",  Z  =  24°  31'  36".  6.  211.40  feet. 
6.    92.685  million  miles. 


Art.  91,  Page  82 

1.  A  =  61°  6'  29",  B  =  58°  46'  4",  C  =  60°  7'  24".  2.  A  =  132°  45' 4", 
B  =  10°  28'  36",  C  =  36°  46'  26".  3.  X  =  35°  39'  50",  Y  =  81°  56'  20", 
Z  =  62°  23'  50".  4.  A  =  44°  24'  50",  B  =  57°  7'  20",  C  =  78°  27'  50". 
5.  Angles  should  be  85°  50'  24",  68°  7'  4",  26°  2' 28".  6.  N  42J  50'  40"  E 
or  S  42°  50'  40"  E. 


Art.  94,  Page  86 

5.  A  =  .055898,  ff  =  .3033.  6.  A  =  3.3797,  R  =  1.6544.  7.  A  =  697.4, 
R  =  35.648.  8.  A  =  90,218,  R  =  316.89.  9.  A  =  426.6,  -  =  S.6172. 
10.     A  =  .16379,   r  =  .15166.       11.     A  =  157,375,    r  =  128.71.       12.      A  = 

.00 16056,  r  =  .016573. 


108  ANSWERS 

Art.  96,  Pages  86-89 

1.  1,387.1  yards,  1,106.7  yards,  121°  16' 7",  15°  43' 53".  2.  5.1818 
miles;  16.6  min.  3.  2,722.7  feet.  4.  71.393  feet.  5.  c  =  75.527,  B  = 
20°  38' 55",  C  =  131°  35'  49",  A  =  626.3.  6.  A  =  21°  47'  12",  B  =  38° 
12'  46",  C  =  120°,  A  =  6.4951.  7.  b  =  23.962,  A  =  65°  47'  0",  £  =  71° 
50' 0",  A  =  185.75;  6'  =  10.016,  A'  =  114°  13' 0",  B'  =  23°  24' 0",  A'  = 
77.642.  8.  A  =  7,768.8,  r  =  28.433,  R  =  144.33.  9.  A  =  1.5928,  r  = 
.54269,  R  =  1.1556.  10.  A  =  124.44,  r  =  4.3665,  R  =  12.807.  11.  36° 
17'  58",  108°  42'  2".  12.  239.93  feet,  129.53  feet.  13.  300.92  feet.  14. 
106.84  miles.  15.  N  75°  E.  16.  1,849.6  feet.  17.  109.34  feet.  18.  16.07 
feet,  346.40  feet.  19.  96.742  feet,  28.258  feet,  133°  50' 14",  12°  9' 46". 
20.  15  Vf  feet.  21.  V23  feet.  22.  61.699  feet.  23.  266.88  feet.  24, 
19.04  feet.     25.   84.01  feet. 


Art.  98,  Page  94 
1.    u  «  (t  +  3)/4.     2.    u  =  V3l.     3.    u  =  10.     4.    u  =  </i 

Art.  101,  Page  97 
1.    ±60°,  ±420°,  etc.;  45°,  225°,  etc.;  30°;  0°;  0°;  180°;  etc.;  45°.    2.    1/3; 
2/5;  Vf.   3.    ±V3/2;  ±1;  4;  ±.9018;  1;  I/V3.     4.   0;  0;  (V3  +  i)/2  VJF. 


Art.  106,  Page  103 

1.  45°  +  n  •  180°,  71°  33'  54"  -f  n  ■  180°.  2.  nir  +  (-1)  nn/6  -tt/4. 
3.  n-  180°  +  (-1)"  41°  48'  39".  4.  nw  +(-l)n  +  17r/6,  rnr  +  (-1)  n  +  1  tt/2. 
5.   2n;r,      (2  n  + l)7r/2.  6.   nx±  (-l)Mx/6.  7.   n- 180* +  49° 

36'34",n-  180° -23°  2' 39".  8.  n  ■  360°  ±  51°  49'  30".  9.  n-  180°  +  (-l)r' 
38°  10' 30".  10.  nx,  2  nx  ±  x/3.  2  «tt  ±  2  tt/3.  11.  Base  angle  =  61° 
55' 40";  base  =  6.4.  12.  Sides  are  15  and  9;  angles  are  36°  52'  12"  and 
53°  7'  48".  13.  Chords  are  8  and  6.  14.  A  =  60°,  a  =  b  =  10.  15.  97.42 
and  22.58  pounds.     16.   b  =  16.7964,  c  =  23.204,  C  =  77°  39'. 


INDEX 


(The  Numbers  refer  to  Articles) 


Abscissa,  3 
Acute  angle,  8 
Angle,  8,  48 

of  depression,  48 

of  elevation,  48 
Area  of  a  triangle,  76,  92 
Axis,  2 

of  abscissae,  3 

of  coordinates,  3 

of  ordinates,  3 

Base  of  a  power,  29 

of  a  system  of  logarithms,  31 
Bearing,  48 

Characteristic  of  logarithms,  35 
Checking  calculations,  42,  83 
Circle,  92 

Circumscribed  circle,  92 
Circular  measure,  10 
Clockwise  rotation.  8 
Common  logarithms,  35 
Compass,  48 

Complimentary  angles,  8 
Co-ratio,  14 
Cosecant,  14,  62 
Cosine,  14,  51,  81 
Cotangent,  14,  57 
Counterclockwise  rotation,  8 

Degree,  8 

Dependent  variable,  96 
Depression,  angle  of,  48 
Directed  angle,  8 

Elevation,  angle  of,  48 
Equations,  trigonometric,  103,  104 
Exponent,  29 


Fractional  exponent,  29 
Function,  even,  55 

inverse,  96,  97 

multiple-valued,  96 

odd,  55 

single-valued,  96 

trigonometric,  24 

Grade,  10 

Graphing,  50 

Graphs,  51,  57,  62,  97,  100 

Independent  variable,  96 
Initial  side,  8 
Inscribed  circle,  92 
Inverse  cosecant,  100 

cosine,  100 

cotangent,  100 

function,  96 

secant,  100 

sine,  99 

tangent,  100 

Law  of  cosines,  81 
Laws  of  exponents,  29 

logarithms,  33 
Law  of  sines,  76    — 
Law  of  tangents,  84 
Logarithm,  30 

Mantissa  of  a  logarithm,  35 
Mollweide's  equation,  85 

Negative  exponents,  29 

Obtuse  angle,  8 
Ordinate,  3 
Origin,  2 

of  coordinates,  3 


109 


110 


INDEX 


Period,  26 
Periodicity,  26 
Power,  29 

Principal  value,  100 
Projection,  6,  45 

Quadrant,  3 

Radian,  10 
Radius  vector,  14 
Right  angle,  8 
Rotation,  clockwise,  8 
counterclockwise,  8 

Secant,  14,  62 
Sine,  14,  51,  76 
Standard  position,  12 
Straight  angle,  8 


Submultiple,  10 
Supplementary  angles,  8 

Tangent,  14,  57,  84 
Terminal  side,  8 

Trigonometric  equations,  103,  104 
Trigonometric  functions,  24 
ratios,  14 

Unit  distance,  2 

Variable,  96 

x-coordinate,  3 
X-axis.  3 

y-coordinate,  3 
Y-axis,  3 

Zero-th  power,  20 


r 


3   1205  00260  0144 


UC  SOUTHERN  REGIONAL  LIBRARY  FACILITY 


A  A      000  084128 


